ELECTRICAL 
ENGINEERING 


THE  THEORY  AND  CHARACTERISTICS 

OF  ELECTRICAL  CIRCUITS 

AND  MACHINERY 


BY 
CLARENCE  V.  CHRISTIE,  M.  A.,  B.  Sc. 

ASSOCIATE  PROFESSOR  OP   ELECTRICAL   ENGINEERING,    MCGILL 
UNIVERSITY,    MONTREAL,    CANADA 


SECOND  EDITION 
REVISED  AND  ENLARGED 


SECOND  IMPRESSION 


McGRAW-HILL  BOOK  COMPANY,  INC. 
239  WEST  39TH  STREET.     NEW  YORK 


LONDON:  HILL  PUBLISHING  CO.,  LTD. 

6  &  8  BOUVERIE  ST.,  E.  C. 

1917 


'?/:/ 

Engineering 
Library 


COPYRIGHT,  1913,  1917,  BY  THE 
McGRAw-HiLL  BOOK  COMPANY,  INC. 


n  j  C 


THE  MAPLE  PRESS  YORK  PA 


PREFACE  TO  THE  SECOND  EDITION 

THE  second  edition  contains  all  the  material  in  the  original  text 
but  much  of  it  has  been  rewritten  and  a  great  deal  of  new  material 
added. 

The  more  important  additions  include  sections  on  complex 
alternating  waves  and  wave  analysis,  on  polyphase  alternating 
current  circuits,  on  the  construction  of  the  characteristic  curves 
of  direct-current  generators  and  motors,  on  the  design  of  direct- 
and  alternating-current  machinery,  on  the  Blondel  diagram  for 
the  synchronous  motor,  on  the  symbolic  method  of  analysis  of 
the  induction  motor,  on  alternating-current  commutator  motors, 
and  finally  a  chapter  on  electrical  measuring  instruments. 

The  chapter  on  direct-current  machinery  has  been  entirely 
rewritten  and  much  enlarged,  and  to  make  it  complete  a  short 
chapter  outlining  the  design  of  a  direct-current  generator  has 
been  added. 

This  chapter  and  the  other  sections  dealing  with  design  are  not 
intended  to  cover  the  work  required  in  a  course  on  design  but  only 
to  give  the  student  some  idea  of  the  formulae  and  constants 
involved. 

The  chapter  on  measuring  instruments  has  been  placed  at  the 
end  of  the  book  because  some  of  the  principles  involved  cannot 
be  well  understood  by  the  student  until  he  has  mastered  the 
theory  of  the  more  important  electrical  machines. 

It  was  originally  intended  to  introduce  problems  at  the  end  of 
each  chapter  to  be  worked  out  by  the  student  and  also  to  give  a 
list  of  references  to  the  most  important  articles  covering  the  sub- 
ject matter  of  the  chapter,  but  there  was  danger  of  making  the 
volume  too  cumbersome.  A  number  of  good  books  of  problems 
have  been  published  recently  and  the  student  will  find  good  lists 
of  references  at  the  ends  of  the  various  sections  in  the  Standard 
Handbook. 

My  thanks  are  due  to  those  teachers  who  have  offered  valuable 
criticisms  of  the  arrangement  and  method  of  treatment  in  the 
first  edition.  ,  *• 

CLARENCE  V.  CHRISTIE. 

McGiLL  UNIVERSITY,  MONTREAL, 
August,  1917. 


PREFACE  TO  THE  FIRST  EDITION 

THIS  book  has  been  compiled  as  a  foundation  for  lecture  courses 
for  junior  and  senior  students  in  Electrical  Engineering. 

The  theory  and  characteristics  of  electrical  machines  are 
developed  from  the  fundamental  principles  of  electrostatics  and 
electromagnetics.  Only  the  more  standard  types  have  been  dis- 
cussed since  familiarity  with  the  principles  of  .their  operations  will 
guide  the  student  to  a  complete  understanding  of  other  machines 
which  differ  only  in  minor  respects.  This  general  groundwork 
may  be  extended  to  suit  the  requirements  of  particular  classes. 

C.  V.  C. 

McGiLL  UNIVERSITY,  MONTREAL, 
October  14,  1913. 


Vll 


CONTENTS 

PA.OE 

PREFACE   v 

CHAPTER  I 
ELECTROSTATICS 

1.  Electrification 1 

2.  Electrical  conductors  and  insulators 1 

3.  Electrostatics  and  electromagnetics 2 

4.  Laws  of  electrostatics 2 

5.  Coulomb 2 

6.  Electrostatic  field 3 

7.  Field  surrounding  a  point  charge 3 

8.  Dielectric  flux  from  a  unit  charge 4 

9.  Field  between  two  point  charges 4 

10.  Field  between  parallel  plates 5 

11.  Potential 5 

12.  Volt 6 

13.  Induced  charges  of  electricity 6 

14.  Equivalent  charges 7 

15.  Distribution  of  potential  in  the  space  surrounding  a  point  charge    .  7 

16.  Potential  at  a  point  due  to  a  number  of  charges 8 

17.  Potential  at  a  point  due  to  a  charged  surface 8 

18.  Equipotential  surfaces 9 

19.  Potential  gradient 9 

20.  Potential  and  potential  gradient  near  a  charged  sphere 10 

21.  Potential  and  potential  gradient  between  two  charged  spheres     .    .  11 

22.  Capacity 12 

23.  Farad 13 

24.  Dielectric  permeance 13 

25.  Condenser 14 

26.  Parallel  plate  condenser 14 

27.  Capacity  of  concentric  cylinders 15 

28.  Potential  and  potential  gradient  between  concentric  cylinders  .    .  16 

29.  Capacity  of  parallel  conductors 17 

30.  Potential  and  potential  gradient  between  parallel  conductors    .    .  19 

31.  Dielectric  field  and  equipotential  surfaces  between  parallel  con- 

ductors   21 

32.  Capacity  of  a  single  wire  to  earth s 22 

33.  Capacity  of  a  sphere  to  earth 23 

34.  Condensers  in  multiple 24 

35.  Condensers  in  series 24 

36.  Energy  stored  in  a  condenser 25 

ix 


x  CONTENTS 

i 

PAGE 

37.  Stresses  in  an  electrostatic  field 27 

38.  Force  exerted  on  a  dielectric  by  an  electrostatic  field 28 

39.  Effects   of  introducing  dielectrics  of  various  specific  inductive 

capacities  into  a  uniform  field 30 

40.  Effect  of  introducing  conductors  into  electrostatic  fields     ....  34 

41.  Graded  insulation  for  cables 35 

42.  Air  films  in  generator  slot  insulation  .    .    . 39 

43.  Condenser  bushing 40 

44.  Dielectric  strength 41 

45.  Breakdown 43 

46.  Dielectric  losses 44 

47.  Surface  leakage 45 

48.  Corona 45 

CHAPTER  II 

MAGNETISM  AND  ELECTROMAGNETICS 

49.  Magnetization 46 

50.  Laws  of  magnetism 47 

51.  Magnetic  field 47 

52.  Magnetic  flux 47 

53.  Flux  from  unit  pole 48 

54.  Magnetic  potential 48 

55.  Magnetomotive  force • 49 

56.  Permeability 50 

57.  Magnetic  reluctance 50 

58.  Permeance 51 

59.  Electromagnetics 51 

60.  Laws  of  induction 52 

61.  Unit  of  electromotive  force 52 

62.  Force  exerted  by  a  magnetic  field  on  an  electric  circuit 53 

63.  Unit  current 54 

64.  Transformation  of  mechanical  energy  to  electrical  energy  ....  54 

65.  Electric  power  and  energy 55 

66.  Intensity  of  magnetic  fields  produced  by  electric  currents ....  56 

67.  Magnetomotive  force  of  a  solenoid 61 

68.  Examples 61 

69.  Energy  stored  in  the  magnetic  field .,   .  >    . '.    .  66 

70.  Stress  in  the  magnetic  field -., 67 

71.  Force  between  parallel  wires  carrying  current 68 

72.  Magnetic  characteristics 69 

73.  Hysteresis .   ...    . 71 

74.  Magnetic  materials 73 

75.  Effect  of  chemical  composition  and  physical  treatment  on  hystere- 

sis loss 74 

76.  Theories  of  magnetism 75 

77.  Lifting  magnets 76 


CONTENTS  xi 

CHAPTER  III 

ELECTRIC  CIRCUITS 

PAGE 

78.  Ohm's  Law 79 

79.  Joule's  law  . 79 

80.  Heat  units 80 

81.  Examples 80 

82.  Resistance 80 

83.  Conductance 81 

84.  Effect  of  temperature  on  resistance 81 

85.  Properties  of  conductors 83 

86.  Resistance  of  conductors 84 

87.  Drop  of  voltage  and  loss  of  power  in  a  distributing  circuit    ...  86 

88.  Current-carrying  capacity  of  wires 87 

89.  Examples 87 

90.  Kirchoff's  laws 88 

91.  Examples 89 

92.  Resistances  in  series 90 

93.  Resistances  in  parallel 90 

94.  Potentiometer 91 

95.  Inductance 92 

96.  Examples 93 

97.  Inductance  of  circuit^  containing  iron 94 

98.  Mutual  inductance  and  self -inductance 95 

99.  Self-inductance  of  continuous-current  circuits 97 

100.  Example 101 

101.  Inductance  of  parallel  conductors 102 

CHAPTER  IV 

ELECTRIC  CIRCUITS  (CONTINUED) 

102.  The  sine  wave  of  electromotive  force  and  current 104 

103.  The  average  value  of  a  sine  wave 106 

104.  The  effective  value  of  a  sine  wave 106 

105.  Inductance  in  alternating-current  circuits      . 107 

106.  Resistance  and  reactance  in  series 109 

107.  Capacity  in  alternating-current  circuits Ill 

10&.  Resistance  and  condensive  reactance  in  series 112 

109.  Resistance,  inductance  and  capacity  in  series 113 

110.  Vector  representation  of  harmonic  quantities 115 

111.  Power  and  power  factor 116 

112.  Examples 122 

113.  Numerical  examples 126 

114.  Circuit  constants •....: 131 

115.  Rectangular  coordinates 134 

116.  Examples  in  rectangular  coordinates .  138 

117.  Kirchoff's  laws  applied  to  alternating-current  circuits 141 


xii  CONTENTS 

CHAPTER  V 

COMPLEX  ALTERNATING-CURRENT  WAVES 

PAGE 

118.  Complex  alternating  waves 142 

119.  Examples 143 

120.  Analysis  of  alternating  waves 146 

121.  Example  of  analysis 149 

CHAPTER  VI 

POLYPHASE  ALTERNATING-CURRENT  CIRCUITS 

122.  Polyphase  Alternating-current  circuits 152 

123.  Three-phase  circuits       154 

124.  Electromotive  forces,  currents  and  power  in  three-phase  circuits  156 

125.  Measurement  of  power  in  polyphase  circuits 159 

126.  Examples 164 

CHAPTER  VII 

DIRECT-CURRENT  MACHINERY 

127.  The  direct-current  dynamo : 165 

128.  Yoke 165 

129.  Pole  pieces  .    .    .    : 165 

130.  Armature  core , 167 

131.  Armature  winding      167 

132.  Ring  windings 167 

133.  Drum  winding • 168 

134.  Multiple-drum  windings 171 

135.  Equalizer  rings 171 

136.  Series-drum  windings 172 

137.  Double  windings 174 

138.  Commutator 176 

139.  Brushes  and  brush  holders 178 

140.  Field  windings 178 

141.  Direction  of  rotation  of  generators  and  motors 179 

142.  Generation  of  electromotive  force 180 

143.  Effect  of  moving  the  brushes 181 

144.  Building  up  of  electromotive  force  in  a  self -excited  generator    .    .    182 

145.  Armature  reaction  and  distribution  of  magnetic  flux       182 

146.  No-load  saturation  curve 186 

147.  Load  saturation  curves 187 

148.  Voltage  characteristic  or  regulation  curve     .    .    .    .  >    .    .    .    .    .   188 

149.  Field  characteristic  or  compounding  curve 190 

150.  Voltage  characteristic  of  a  shunt  generator .   191 

151.  Effect  of  change  of  speed  on  the  voltage  of  a  shunt  generator  .    .192 

152.  Effectof  saturation  on  the  voltage  characteristic  of  a  shunt  generator.  193 


CONTENTS  xiii 

PAGE 

153.  Compound  generator 194 

154.  Voltage  characteristic  of  a  compound  generator 195 

155.  Short-shunt  and  long-shunt  connection      196 

156.  Regulation .......  196 

157.  Series  generator .    ....    .    ...    .    .    .  196 

158.  Electric  motors   .    .    .    .    . 198 

159.  Types  of  motors      . 198 

160.  Speed  equation  of  a  motor 199 

161.  Methods  of  varying  speed \    ......  199 

162.  Speed  characteristics  of  motors 201 

163.  Torque  equation .    .   v    .    .    ...            .    .  203 

164.  Torque  characteristics  of  motors .  204 

165.  Construction  of  the  speed  characteristics 204 

166.  Construction  of  the  speed  characteristic  for  a  series  motor    .    .    .  205 

167.  Variation  of  speed  of  a  shunt  motor  with  line  voltage 206 

168.  Variation  of  speed  with  temperature  of  the  field  coils 207 

169.  Construction  of  the  torque  characteristics 207 

170.  Starting  torque 208 

171.  Motor  starter  with  no-voltage  release 209 

172.  Adjustable  speed  operation 210 

173.  Multiple-wire  systems  of  speed  control 211 

174.  Ward  Leonard  system  of  speed  control 211 

175.  Speed  control  of  series  motor 212 

176.  Interpole  motors «    .*....... 213 

177.  Applications  of  motors  . 213 

178.  Losses  in  direct-current  machinery 215 

179.  Shunt-field  loss !    .  ....  f 215 

180.  Series-field  loss 215 

181.  Armature  copper  loss 216 

182.  Loss  at  the  brush  contacts «    *i  .    .    -  216 

183.  Hysteresis  loss    ......    .    .    .    .    .    ,    .    .    .    .    .    .    .    ,    .    .  217 

184.  Eddy  current  loss 218 

185.  Pole-face  loss       .    .    ,    .    .....    .    .    .    .    ,    .    .    ....    ...  219 

186.  Brush-friction  loss      v    .........    .  220 

187..  Bearing-friction  and  windage  losses .    ;    ;..    .  220 

188.  Eddy-current  losses  in  the  armature  copper  .    .    .    ...    .    .    .    .  221 

189.  Constant  losses  and  variable  losses 222 

190.  Core  loss.    .    .    .    .    ...    .    .    .    .    ,   .    .'. 222 

191.  Efficiency.    .    .    .    .    .    ,    .    .    .    <   .    .-;:—. 222 

192.  Commutation 224 

193.  Commutating  electromotive  force .    .    .  231 

194.  Conditions  essential  to  sparkless  commutation 233 

195.  Calculation   of  the  reactance  voltage  for  a  full-pitch  multiple 

winding .    ...._.    .    .    .  V  ,  -,    .    .    .  233 

196.  Current  density  at  the  brush  contacts    .    .    .    .    .    .    .    .    .-. ,..    .  236 

197.  Burning  of  the  brush  and  commutator 236 

198.  Poor  commutation  resulting  from  too  many  coil  sides  per  slot  .    .  237 

199.  Interpoles 237 


xiv  CONTENTS 

PAGE 

200.  Compensating  windings     .    . ..    ...    4    ;  ".    .  ,.  .•„    .  239 

201.  Flashing .    ....    >   v-  .    .>.  ;.    .    .";*•....:„.'.  240 

202.  Parallel,  operation .    .   v  V- :  ••.i_. -v.'v. ;..'..  242 

203.  Parallel  operation  of  compound  generators    .....   ..  .;;  -  .    .    .    .  243 

204.  Storage  batteries ...    .^  ..:;.-.....  245 

205.  Applications « :.-••;  - . .  .  •  .    .    .  246 

206.  Boosters ...    ..:....  247 

207.  Balancers .    V   .    .  249 

208.  Rosenberg  generator  for  train  lighting •...-..  250 

209.  Homopolar  generators 251 

210.  Limits  of  output  of  electric  machines 252 

211.  Temperature  limits    .    .    .    . ...    .    .    .  253 

212.  Temperature  of  commutators 255 

213.  Temperature  of  cores     .    .    <' .    . .   ..    .    ,;  .    .  255 

214.  Temperature  of  other  parts      ...    t ...    .    .    .  255 

215.  Ventilation 256 

216.  Semi-enclosed  and  totally  enclosed  machines    .    .    .    ......    .  256 

CHAPTER  VIII 

DESIGN  OF  A  DIRECT-CURRENT  GENERATOR 

217.  Symbols 258 

218.  Design  of  direct-current  machinery 259 

219.  Magnetic  leakage ;,    ...  263 

220.  Design  of  a  direct-current  generator ..;...  264 

221.  Flux  in  the  air  spaces  between  the  teeth .    .  267 

222.  Effect  of  tooth  taper ,    .    .    .  268 

223.  Ampere-turns  per  inch  for  an  air  path .  -.,;.    .    •  268 

224.  Air  gap ^    .    ..:X,,.    .  269 

225.  Length  of  the  air  gap •    •    V  •    •  269 

226.  Design  of  poles  and  yoke 270 

227.  Determination  of  the  no-load  saturation  curve      .    .    :    .    ,    .    .  270 

228.  Field  winding •    -.    •    •    -  273 

229.  Current  density  in  field  windings .    .    .    .,  ».  /.•    -.274 

230.  Size  of  wire  for  the  field  winding -.  >    ;  ;:.;.  ^,...    .    .  274 

231.  Resistance  of  the  armature  winding    .    .    . 275 

232.  Determination  of  the  magnetomotive  force  of  the  series  winding  275 

233.  Design  of  the  series-field  winding 277 

234.  Design  of  the  commutator    .    . •    •    •  278 

235.  Losses ,  ,  ....    .V>-  .'V .  279 

CHAPTER  IX 
SYNCHRONOUS  MACHINERY 

236.  Alternator 280 

237.  Types  of  alternators 280 


CONTENTS  xv 

PAGE 

238.  Electromotive  force  equation 284 

239.  Form  factor 286 

240.  Polyphase  alternating-current  generators .  286 

241.  Alternator  windings " .'....  289 

242.  Distribution  factors 290 

243.  Multiple-circuit  windings 294 

244.  Short-pitch  windings 294 

245.  Effect  of  distributing  the  winding 297 

246.  Harmonics  due  to  the  teeth 297 

247.  Effect  of  third  harmonics  in  three-phase  alternators 299 

248.  General  electromotive  force  equation 300 

249.  Rating  of  alternators.    .    .    .    S •  *    .  300 

250.  Comparative   ratings   of  an  alternator  wound  single,  two-  and 

three-phase      '  301 

251.  Armature  reaction 302 

252.  Armature  reactance 304 

253.  Leakage  fluxes 304 

254.  Polyphase  armature  reaction 305 

255.  Single-phase  armature  reaction 309 

256.  Electromotive  forces  in  the  alternator 311 

257.  Armature  resistance 312 

258.  Vector    diagram    of    electromotive   forces    and    magnetomotive 

forces 313 

259.  Voltage  characteristics 315 

260.  Compounding  curves 317 

261.  Tests   for   the   determination    of  the  regulation  of  alternating- 

current  generators 317 

262.  Regulation 321 

263.  Short-circuit  of  alternators 323 

264.  Synchronous  motor 327 

265.  Vector  diagrams  for  a  synchronous  motor 328 

266.  Characteristic  curves 330 

267.  Compounding  curves 330 

268.  Load  characteristics 331 

269.  Phase  characteristics 333 

270.  Torque 334 

271.  Blondel  diagram  for  a  synchronous  motor 334 

272.  Synchronizing  power 338 

273.  Construction  of  the  characteristic  curves  of  a  synchronous  motor 

from  the  Blondel  diagram 339 

274.  Load  characteristics 340 

275.  Phase  characteristics  or  "V"  curves  and  compounding  curves.    .  341 

276.  Starting  synchronous  motors 342 

277.  Self -starting  motors 343 

278.  Synchronous  phase  modifier 345 

279.  Parallel  operation  of  alternators      .,,',..  346 

280.  Effect  of  inequality  of  terminal  voltage 346 

281.  Effect  of  inequality  of  frequency 347 


xvi  CONTENTS 

PAGE 

282.  Effect  of  difference  of  wave  form 350 

283.  Conclusions 351 

284.  Hunting .  .."• 351 

285 .  Frequency  of  hunting    .    .    . 352 

286.  Design  of  alternating-current  generators  and  motors 354 

287.  Electromotive  force  equation 354 

288.  Output  equation 355 

289.  Flux  densities 356 

290.  Current  densities 356 

291.  Insulation  for  high-voltage  alternators 356 

292.  Extra  insulation  required  under  special  conditions 357 

293.  Armature  windings 357 

294.  Slots  per  pole 357 

295.  Regulation 357 

296.  Excitation  regulation 358 

297.  Excitation .358 

298.  Losses 359 

299.  Ventilation 359 

300.  Cylindrical  rotors 361 

CHAPTER  X 

TRANSFORMERS 

301.  The  constant-potential  transformer 363 

302.  Vector  diagrams  for  the  transformer 366 

303.  Exciting  current 368 

304.  Leakage  reactances 368 

305.  The  transformer  as  a  circuit 370 

306.  Examples ;    .    .   371 

307.  Measurement  of  the  constants  of  a  transformer 374 

308.  Regulation ,    ,       376 

309.  Voltage  characteristics 378 

310.  Losses  in  transformers       378 

311.  Hysteresis  loss ,   v.,    ...   378 

312.  Eddy-current  loss  in  transformer 379 

313.  Extra  losses 380 

314.  Efficiency 381 

315.  Types  of  transformer 383 

316.  Methods  of  cooling ,.'......,. 384 

317.  Transformer  connections .    .    .   386 

318.  Single-phase  transformers  on  polyphase  circuits 387 

319.  Star-star  connection  .    . 388 

320.  Delta-star  and  star-delta  connection  .    . 390 

321.  Delta-delta  connection .        .   391 

322.  Open-delta  connection ' 391 

323.  "T"  connection .    .   392 

324.  Transformation  from  two-phase  to  three-phase 393 


CONTENTS  xvii 

PAGE 

325.  Single-phase  power  from  three-phase  circuits 395 

326.  Multiple  operation  of  transformers 395 

327.  Booster  transformers 39(5 

328.  Auto-transformers.    .•'«•«•. 396 

329.  Instrument  transformers 397 

330.  Current  transformers. '. ...  399 

331.  Example  .............  ^ 401 

332.  The  constant-current  transformer 401 

333.  Regulation 402 

334.  Power  factor    .    ...    .    ...    .    . 403 

335.  Induction  regulator 403 

336.  Design  of  a  transformer 405 

337.  Reactance 408 

338.  Design  of  a  shell-type,  water-cooled  transformer,  single-phase,  60 

cycles,  1,000  kva.,  22,000  to  2,200  volts  for  power  transmission .  4.10 

CHAPTER  XI 
CONVERTERS 

339.  Types  of  converters 414 

340.  Synchronous  converter 414 

341.  Ratios  of  e.m.fs.  and  currents 415 

342.  Two-phase  or  quarter-phase  converter 417 

343.  Three-phase  converter 418 

344.  n-phase  converter  .    .    .    .    .    .    . .    .    .    .    .  419 

345.  Wave  forms  of  currents  in  the  armature  coils  .    .'.'.'.    .    .    .    .  421 

346.  Heating  due  to  armature  copper  loss  .    . ' 422 

347.  Armature  reaction  .    .    . 426 

348.  Control  of  the  direct-current  voltage '.    .    ,    .    .    .    .  429 

349.  Methods  of  varying  the  alternating  voltage      ...:.....  430 

350.  Compounding  by  reactance • ;    .    .  430 

351.  Synchronous  booster  converter «/i    .    .    .    .    .431 

352.  Direct-current  booster  converter .    .x 431 

353.  Split-pole  converter    .    .    .' '. .  432 

354.  Frequencies  and  voltages  .    .    ....    .    .    ...    ......    .    .  432 

355.  Outputs  and  efficiencies •-.<•'. :  .    .    .    .  433 

356.  Overload  capacity 433 

357.  Dampers ,    .    .-.^.    .    .   .    .    .    .    .  433 

358.  Starting    .    .    .....    ..... .  433 

359.  Alternating-current  self-starting 433 

360.  Direct-current  self-starting 434 

361.  Starting  by  an  auxiliary  motor 434 

362.  Brush  lifting  device 434 

363.  Bucking  and  flashing 435 

364.  Parallel  operation 435 

365.  Inverted  converter 435 

366.  Double-current  generator.    ...'.. 436 


xviii  CONTENTS 

J*AGE 

367.  Three-wire  generator 437 

368.  Parallel  operation  of  three-wire  generators 439 

369.  Frequency  converters i.  .  439 

370.  Mercury  vapor  rectifier 440 

371.  Currents  and  voltages ; v  ......  441 

372.  Losses  and  efficiency : 442 

373.  Three-phase  rectifier 442 

374.  Hot-cathode  argon-filled  rectifier 442 

CHAPTER  XII 
INDUCTION  MOTOR 

375.  Induction  motor 444 

376.  The  stator 444 

377.  Revolving  magnetomotive  force  and  flux  of  the  stator 447 

378.  The  rotor 450 

379.  Slip 450 

380.  Magnetomotive  force  of  the  rotor 451 

381.  Electromotive  force  and  flux  diagram  for  the  induction  motor .    .  451 

382.  Proof  that  the  locus  is  a  circle 454 

383.  Magnetomotive  force  diagram =  •.-.-...    .  454 

384.  Stator  current  diagram 454 

385.  Rotor  electromotive  force  and  current ' .    . •  .    .  456 

386.  Rotor  input 456 

387.  Rotor  copper  loss  and  slip 457 

388.  Rotor  output       457 

389.  Torque : .    .    .        ...  457 

390.  Rotor  efficiency 458 

391.  Modification  of  diagram    .    .    .   < 458 

392.  Interpretation  of  diagram 460 

393.  Construction  of  diagram  from  test  for  a  three-phase  motor   .    .    .461 
494.  Test  of  an  induction  motor 462 

395.  Analysis  by  rectangular  coordinates 465 

396.  Characteristics  of  an  induction  motor  by  the  symbolic  method     .  470 

397.  Methods  of  starting 473 

398.  Applications «.   .........  475 

399.  Speed  control  of  induction  motors  .    .    .    .    .    .-;.-. 476 

400.  Voltage  control .    .  ".  :. 476 

401.  Rotor  resistance  control 476 

402.  Pole-changing ',."*-• 476 

403.  Cascade  control  or  concatenation    .    .    .    .,;,.. 476 

404.  Exciting  current 478 

405.  Leakage  reactances ,    .  '.  j.    .  479 

406.  Stator  and  rotor  resistances 480 

407.  Effect  of  change  of  voltage  and  frequency    .    .    .    .    .    .   V -.    .    .  481 

408.  Single-phase  induction  motor •-«.. 482 

409.  Horizontal  field  at  slips >,......,.  484 


CONTENTS  xix 

PAGE 

410.  Starting  single-phase  induction  motors 484 

411.  Comparison  of  single-phase  and  polyphase  motors 485 

412.  Induction  generator  .    .    . 486 

413.  Asynchronous  phase  modifier 488 

414.  Phase  converter 491 

415.  Induction  frequency  converter .'....    ...    .    .  493 

CHAPTER  XIII 
ALTERNATING-CURRENT  COMMUTATOR  MOTORS 

416.  Motor  characteristics 494 

417.  Alternating-current  series  motor •;. 494 

418.  Design  for  minimum  reactance 497 

419.  Compensating  windings 497 

420.  Commutation .....  .    .   ...  .    ...    .    ...    .    .    .  499 

421.  Speed  control 500 

422.  Polyphase  commutator  motor 501 

423.  Repulsion  motor ...    ,    .    ...........  501 

424.  Commutation .    .'.    ...    .    .    .    .    .    .    .    .    .  503 

425.  Compensated  repulsion  motor      '.    .    .....    .    , 504 

426.  Alternating-current  commutator  motors  with  shunt  characteristics  505 

427.  Single-phase  induction  motor  with  repulsion  starting 506 

CHAPTER  XIV 
TRANSMISSION  SYSTEMS 

428.  Transmission  line 507 

429.  Relative  amounts  of  conducting  material  for  single-,  two-  and 

three-phase  transmission  lines 507 

430.  Reactance 508 

431.  Capacity  ......... 509 

432.  Voltage  and  frequency . ....    .  510 

433.  Spacing  of  conductors .    .  510 

434.  Single-phase  transmission  line  .    .    . .    . .'.  510 

435.  Three-phase  transmisson  line . 517 

436.  Application  of  a  synchronous  phase  modifier  to  a  transmission 

system ..........  521 

437.  High-voltage  direct-current  system ....    .    .    .  524 

438.  Advantages  and  disadvantages  of  the  Thury  or  series  system   .    .  525 

CHAPTER  XV 
ELECTRICAL  INSTRUMENTS 

439.  Electrical  instruments 527 

440.  Direct-current  voltmeters  and  ammeters 527 

441.  Voltmeter  multipliers 527 


xx  CONTENTS 

PAGE 

442.  Ammeter  shunts .    .    .    .    .    .  •  *•  v   v   *    , 528 

443.  Thomson  inclined-coil  ammeter 528 

444.  Weston  soft-iron-type  ammeters  and  voltmeters 528 

445.  Electrodynamometer-type  voltmeter 529 

446.  Hot-wire  ammeters  and  voltmeters 529 

447.  Dynamometer-type  wattmeter 530 

448.  Induction-type  wattmeter 531 

449.  Power-factor  meters 532 

450.  Frequency  meters 533 

451.  The  Weston  frequency  meter 534 

452.  Synchroscope 534 

Tirrill  regulator 535 

454.  Automatic  voltage  regulator  for  alternating-current  generators     .  537 

INDEX                                                                                                                .  541 


ELECTRICAL  ENGINEERING 


CHAPTER  I 
ELECTROSTATICS 

1.  Electrification. — Bodies  which  are  charged  with  electricity 
are  said  to  be  electrified.     Charges  are  of  two  kinds  called  posi- 
tive and  negative.     Bodies  which  have  a  positive  charge  are 
acted  upon  by  forces  tending  to  make  them  give  up  their  charge; 
bodies  which  have  a  negative  charge  are  acted  upon  by  forces 
tending  to  convey  a  positive  charge  to  them  equal  to  their  nega- 
tive charge.     These  forces  are  exerted  through  the  medium  sepa- 
rating the  charges  and  the  medium  is  in  a  state  of  stress. 

The  body  with  the  positive  charge  is  at  a  higher  potential  than 
the  body  with  the  negative  charge  and  the  difference  of  potential 
between  the  two  is  a  measure  of  the  tendency  for  electricity  to 
pass  from  one  to  the  other. 

2.  Electrical   Conductors   and   Insulators. — If   two   metallic 
bodies  charged  to  different  potentials  are  joined  by  a  metal  wire, 
electricity  will  flow  from  one  to  the  other  until  the  potential  of 
both  is  the  same  and  the  transfer  of  electricity  will  take  place 
almost  instantaneously.     The  metal  wire  is  therefore  a  good  con- 
ductor of  electricity;  or,  it  offers  a  low  resistance  to  the  passage 
of  electricity  through  it. 

If  the  two  charged  bodies  had  been  joined  by  a  glass  rod,  there 
would  have  been  no  transfer  of  electricity  between  them,  or,  it 
would  have  taken  place  so  slowly  that  it  could  only  be  detected 
by  the  most  delicate  instruments.  Glass  is  therefore  a  very  bad 
conductor;  or,  it  offers  a  very  high  resistance  to  the  passage  of 
electricity.  It  is  called  a  non-conductor  or  insulator. 

As  all  materials  conduct  to  a  certain  extent,  it  is  not  possible 
to  divide  them  absolutely  into  conductors  and  insulators,  but, 
since  the  resistance  of  a  good  insulator  is  many  million  times 
that  of  a  good  conductor,  they  may  be  so  divided  for  practical 
purposes. 

I 


2  ELECTRICAL  ENGINEERING 

In  the  first  class  are  silver,  which  is  the  best  conductor,  copper 
and  other  metals,  graphite,  impure  water  and  solutions  of  salts. 
In  the  second  class  are  air,  which  when  dry  is  an  almost  perfect 
insulator,  glass,  paraffin,  ebonite,  porcelain,  rubber,  shellac,  oils  and 
the  numerous  insulating  compounds  used  in  electrical  engineering. 

3.  Electrostatics  and  Electromagnetics.  —  Electrostatics  com- 
prises phenomena  related  to  electric  charges  at  rest  and  to  the 
stresses  produced  in  the  fields  surrounding  them.     These  phe- 
nomena become  of  great  importance  where  very  large  differences 
of  potential  must  be  provided  for,  as  for  example  in  the  design 
and  operation  of  all  high  voltage  apparatus  and  systems. 

Electromagnetics  comprises  phenomena  related  to  electricity 
in  motion,  that  is,  to  currents  of  electricity  and  the  magnetic 
fields  produced  by  them.  Almost  all  the  problems  to  be  solved 
by  the  electrical  engineer  come  under  this  head. 

4.  Laws  of  Electrostatics.  —  First  Law.  —  Like  charges  of  elec- 
tricity repel  one  another;  unlike  charges  attract  one  another. 

Second  Law.  —  The  force  exerted  between  two  charges  of  elec- 
tricity is  proportional  to  the  product  of  their  strengths  and  is 
inversely  proportional  to  the  square  of  the  distance  between 
them;  it  also  depends  on  the  nature  of  the  medium  separating 
them. 

This  law  can  be  expressed  by  the  formula, 


where  q  and  #1  are  the  charges  of  electricity,  r  is  the  distance 
between  them  in  centimeters,  ft  is  a  constant  depending  on  the 
medium  separating  the  charges  and  is  called  its  specific  inductive 
capacity  or  dielectric  constant.  The  unit  of  quantity  is  so  chosen 
that  the  dielectric  constant  for  air  is  unity;  for  all  other  sub- 
stances it  is  greater  than  unity.  The  table  on  page  43  gives 
the  dielectric  constants  for  some  of  the  most  common  dielectrics. 
/  is  the  force  in  dynes  exerted  between  the  two  charges;  if  the 
charges  are  of  the  same  kind  the  force  between  them  is  a  repulsion 
and  /  is  positive. 

5.  Coulomb.  —  One  electrostatic  unit  of  quantity  is  that  quan- 
tity which,  when  placed  at  a  distance  of  1  cm.  in  air  from  a 
similar  quantity,  repels  it  with  a  force  of  one  dyne. 

The  practical  unit  of  quantity  is  the  coulomb;  one  coulomb  is 
3  X  109  electrostatic  units;  it  is  the  quantity  of  electricity  which 


ELECTROSTATICS  3 

passes  a  point  in  a  conductor  when  one  ampere  flows  for  one 
second  (see  Art.  63). 

6.  Electrostatic  Field.  —  Any  space  in  which  electrostatic  forces 
act  is  called  an  electrostatic  field.     The  direction  of  the  force  at 
any  point  in  the  field  is  the  direction  in  which  a  unit  positive 
charge  placed  at  the  point  tends  to  move  and  its  intensity  is  the 
force  in  dynes  exerted  on  the  unit  charge. 

The  electrostatic  field  is  conveniently  represented  by  lines  of 
electrostatic  induction  or  dielectric  flux  drawn  in  the  direction  of 
the  force.  In  air  the  number  of  lines  per  square  centimeter  is 
equal  to  the  force  in  dynes  at  the  point  and  in  a  medium  of  di- 
electric constant  k  the  number  of  lines  per  square  centimeter  is 
equal  to  k  times  the  force.  This  may  be  stated  in  another  way: 
Unit  electrostatic  force  produces  one  line  of  dielectric  flux  per 
square  centimeter  in  air  and  k  lines  per  square  centimeter  in  a 
medium  of  dielectric  constant  k. 

The  electrostatic  force  at  a  point  is  expressed  in  dynes  and  is 
represented  by  $;  the  dielectric  flux  density  at  a  point  is  expressed 
in  lines  per  square  centimeter  and  is  represented  by  3D. 

7.  Field  Surrounding  a  Point  Charge.  —  At  a  distance  r  cm.  in 
air  from  an  isolated  charge  q,  a  unit  charge  is  acted  on  by  a  force 

ff  =  |  dynes;  (2) 

and  the  dielectric  flux  density  produced  at  the  point  is 

3)  =  $  —  \  lines  per  square  centimeter.  (3) 

This  density  is  produced  over  the  surface  of  a  sphere  of  radius  r 
and  therefore  the  total  dielectric  flux  from  the  charge  q  is 


imes. 

Consider  the  same  charge  surrounded  by  a  medium  of  dielectric 
constant  k. 

The  force  exerted  on  a  unit  charge  at  a  distance  r  cm.  from  q  is 

= 


the  dielectric  flux  density  produced  is 

£>  =  $k  =  ~  lines  per  square  centimeter, 

and  therefore  the  dielectric  flux  from  charge  q  is,  as  before, 

lines.  (4) 


4  ELECTRICAL  ENGINEERING 

Fig.  1  represents  the  electrostatic  field  about  a  positive  point 
charge.  The  lines  of  flux  extend  out  radially  in  all  directions. 

8.  Dielectric  Flux  from  a  Unit  Charge. — The  dielectric  flux 
from  a  unit  charge  is  4?r  lines  by  equation  4.  Thus  if  a  dielectric 
flux  \f/  starts  from  any  surface  the  positive  charge  on  that  surface 

^ 
is  q  =  -^  units,  and  if  a  flux  ^  ends  on  any  surface  the  negative 

charge  on  that  surface  is  —  q  =  -r-  units. 


-Q 


FIG.  1. — Field  surround-      FIG.  2. — Field  between  two      Fio.  3. — Field  between 
ing  a  point  charge.  point  charges.  two  parallel  plates. 


A  unit  positive  charge  is  always  associated  with  each  4rr  lines 
leaving  a  surface  and  a  unit  negative  charge  with  each  4?r  lines 
entering  a  surface. 

9.  Field  between  Two  Point  Charges. — The  field  between  two 
point  charges  q  and  —q  consists  of  curved  lines  extending  from 
the  positive,  to  the  negative  charge. 

The  direction  and  intensity  of  the  force  at  any  point  such  as 
P  may  be  obtained  by  combining  the  forces  exerted  at  the  point 
by  the  two  charges  as  shown  in  Fig.  2.  If  a  unit  positive  charge 

is  placed  at  the  point  P  it  is  repelled  by  q  with  a  force  /i  =  ~2 


ELECTROSTATICS  5 

and  is  attracted  by  —q  with  a  force  /2  =  ^y     The  resultant 

force  at  the  point  is  9>  and  it  is  the  vector  sum  of  f\  and  /2. 

If  D  cm.  is  the  distance  between  the  charges,  the  force  at  a 
point  on  the  line  joining  the  charges  distant  r  cm.  from  the  charge 
q  is 

^  =  ^  +  (^~p  dynes. 

Since  the  medium  is  air  the  flux  density  at  any  point  is  numeric- 
ally equal  to  the  force  ^  at  the  point. 

10.  Field  between  Parallel  Plates.  —  The  field  between  two 
parallel  plates  A  and  B,  Fig.  3,  charged  respectively  with  +g 
and  —  q  units  per  square  centimeter  consists  of  straight  lines 
normal  to  the  plates  except  near  the  edges  where  they  become 
slightly  curved. 

The  dielectric  flux  density  is  uniform  throughout  the  volume 
between  the  plates,  and  its  value  is 

3D  =  4:irq  lines  per  square  centimeter.  (5) 

If  the  medium  between  the  plates  is  air,  the  electrostatic  force 
at  any  point  is 

ff  =  3D  =  4:irq  dynes.  (6) 

If  the  medium  has  a  dielectric  constant  k,  the  force  is 

£>        4:irq   ,  ,„. 

*  ==  fc  =  "T  ynes*  (  } 

11.  Potential.  —  The  difference  of  potential  between  two  points 
is  measured  by  the  work  done  in  carrying  a  unit  charge  from  one 
to  the  other  against  the  electrostatic  force  in  the  field;  it  is  there- 
fore the  line  integral  of  the  force  between  the  points. 

The  difference  of  potential  between  the  two  plates  in  Fig.  3  is 


f 

Jo 


(8) 


where  t  cm.  is  the  distance  between  the  plates. 

Difference  of  potential  tends  to  cause  electricity  to  flow  from 
one  point  to  the  other  and  is  therefore  called  electromotive  force. 

Unit  difference  of  potential  (electrostatic)  exists  between  two 
points  when  one  erg  of  work  is  done  in  conveying  unit  charge 
from  one  to  the  other. 

The  earth  is  usually  taken  as  the  arbitrary  zero  of  potential, 
and  the  potentials  of  other  bodies  are  given  with  reference  to  it. 


6 


ELECTRICAL  ENGINEERING 


If  a  difference  of  potential  is  produced  between  two  points  on 
a  conductor,  electricity  will  flow  from  the  point  of  high  potential 
until  the  potentials  of  all  parts  of  the  conductor  are  the  same. 

When,  however,  the  difference  of  the  potential  is  maintained 
by  an  electric  generator  a  current  of  electricity  flows  continuously 
from  one  point  to  the  other. 

When  a  difference  of  potential  is  produced  between  two  con- 
ductors separated  by  an  insulating  material,  the  material  is  sub- 
jected to  a  stress  and  lines  of  dielectric  flux  pass  through  it. 

12.  Volt. — The  practical  unit  difference  of  potential  or  electro- 
motive force  is  called  the  volt.     One  electrostatic  unit  is  equal 
to  300  volts. 

13.  Induced    Charges    of    Electricity. — When    a    positively 
charged  body  A,  Fig.  4,  is  brought  near  to  an  insulated  conductor 
BC,  a  negative  charge  is  induced  on  the  nearer  side  B  and  an 
equal  positive  charge  on  the  farther  side  C.     The  explanation  is 


FIG.  4. — Induced  charges. 


FIG.  5. — Induced  charges. 


that  the  potential  of  the  end  B  due  to  the  charge  on  A  is  higher 
than  that  of  C,  but,  since  BC  is  a  conductor,  as  soon  as  there  is 
a  difference  of  potential  between  two  points  on  it,  electricity 
flows  from  the  region  of  high  potential  at  B  to  the  region  of  low 
potential  at  C;  thus  a  positive  charge  appears  at  C  and  an  equal 
negative  charge  at  B  and  these  charges  so  distribute  themselves 
over  the  surface  of  BC  that  the  potential  at  every  point  on  it  is 
the  same,  being  the  sum  of  the  potentials  due  to  the  charge  on  A 
and  the  two  charges  on  BC. 

The  same  result  is  illustrated  in  Fig.  5,  which  shows  the  dielec- 
tric flux  produced. 

The  flux  from  A  extends  out  radially  in  all  directions.  In  the 
region  surrounding  BC  the  lines  are  deflected  and  a  large  number 
pass  through  the  conductor  J5C,  since  its  dielectric  constant  is 
infinite  and  therefore  it  offers  an  easy  path.  For  every  4r  lines 


ELECTROSTATICS  1 

entering  at  B  a  unit  negative  charge  appears  on  the  surface, 
and  for  every  4?r  lines  leaving  at  C  a  unit  positive  charge  appears. 
14.  Equivalent  Charges.  —  A  uniformly  distributed  charge  on 
the  surface  of  a  sphere  acts  as  though  it  were  concentrated  at  the 
center.  If  a  sphere  of  radius  R  cm.  has  a  charge  Q  uniformly 

distributed  over  its  surface,  the  density  of  the  charge  is 


and  since  4?r  lines  emanate  from  unit  charge,  the  flux  density  at 
the  surface  of  the  sphere  is  j~~™  =  ™  lines  per  square  centimeter. 

If  the  charge  Q  is  concentrated  at  the  center  of  the  sphere,  then, 
by  formula  (3),  the  flux  density  at  a  distance  R  cm.  from  the 

charge  is  -™  and  therefore  the  uniformly  distributed  charge  on 

the  surface  of  the  sphere  may  be  represented  by  an  equal  charge 
concentrated  at  the  center. 

Similarly  a  uniformly  distributed  charge  on  the  surface  of  a 
cylinder  may  be  represented  by  an  equal  charge  uniformly  dis- 
tributed along  'the  axis  of  the  cylinder. 

15.  Distribution  of  Potential  in  the  Space  Surrounding  a  Point 
Charge.  —  In  Fig.  6  a  positive  charge  q  is  placed  at  the  point  0 


i  ! 

FIG.  6. — Distribution  of  potential  about  a  point  charge. 

at  an  infinite  distance  from  all  other  charges.  The  potential  at 
a  point  PI,  distant  7*1  cm.  from  0,  is  the  work  done  in  moving 
unit  charge  from  a  point  of  zero  potential  to  the  point  PI  against 
the  forces  in  the  field.  The  intensity  of  the  force  at  a  distance 
r  cm.  from  0  is  by  equation  2 

$  =  ^  dynes; 

the  work  done  in  moving  a  unit  charge  against  this  force  through 
a  distance  dr  is 

dr 


8  ELECTRICAL  ENGINEERING 

and  the  work  done  in  moving  the  charge  from  a  point  of  zero 
potential  to  PI  is 

r«j    r  ^ 

ei  =         5  dr  =    \     q—. 
Jri  Jri  *  r2 

-«H];    ' 

=  ^  ergs;  (9) 

therefore,  the  potential  at  a  point  due  to  a  charge  of  electricity 
is  equal  to  the  charge  in  electrostatic  units  divided  by  its  distance 
in  centimeters  from  the  point. 

16.  Potential  at  a  Point  Due  to  a  Number  of  Charges. — If 
there  are  several  charges  q\,  q%,  q$,  etc.,  in  the  field,  the  potential 
at  any  point  is  the  sum  of  the  potentials  due  to  the  separate 
charges  and  is 


where  ri,  r2,  r3,  etc.,  are  the  distances  from  the  various  charges  to 
the  point  (Fig.  7). 

B 


A 

FIG.  7. — Potential  due  to  a  num-  FIG.  8. — Potential  due  to  a  charged 

ber  of  charges.  surface. 

17.  Potential  at  a  Point  Due  to  a  Charged  Surface. — In  Fig.  8 
AB  is  a  surface  with  a  non-uniform  distribution  of  charge  over  it; 
if  dq  is  a  small  element  of  charge  at  a  distance  r  cm.  from  the 
point  P,  the  potential  at  P  due  to  the  charge  dq  is 

dq 

de  =  —i 

r 

and  the  potential  due  to  the  total  charge  on  the  surface  is 


e  = 


(ID 


ELECTROSTATICS 


9 


18.  Equipotential  Surfaces. — Surfaces  of  which  all  points  are 
at  the  same  potential  are  called  equipotential  surfaces. 

In  Fig.  9,  A  is  an  isolated  sphere  of  radius  R,  charged  with 
Q  units  of  electricity.  Any  spherical  surface  drawn  about  the 
center  of  A  is  an  equipotential  surface.  The  potential  of  surface 

(1)  is  — *  that  of  (2)  is  —  and  that  of  the  sphere  A  is  „-• 

7*1  7*2  it 

The  difference  of  potential  between  surfaces  (1)  and  (2)  is 
and  is  the  work  that  must  be  done  in  taking  a  unit  charge 

from  any  point  on  (2)  to  any  point  on  (1).  It  makes  no  differ- 
ence what  path  the  charge  follows,  because  its  path  can  always 
be  resolved  into  two  displacements,  one 
along  the  equipotential  surface  and  the 
other  normal  to  it;  no  work  is  done  in 
moving  along  the  equipotential  surface, 
since  there  is  no  opposing  force  and  there- 
fore all  the  work  is  done  in  the  displace- 
ment along  the  normal. 

The  electrostatic  force  on  an  equipoten- 
tial surface  is  normal  to  the  surface,  there- 
fore lines  of  induction  pass  normally 
through  equipotential  surfaces. 

Electric  conductors  not  carrying  current  are  equipotential  sur- 
faces, since  if  differences  of  potential  did  exist  electricity  would 
flow  from  the  points  of  high  potential  until  the  potential  became 
uniform  throughout  the  conductor. 

19.  Potential  Gradient. — The  potential  gradient  at  a  point  is 
the  space  rate  of  change  of  potential  at  the  point  measured  in 
the  direction  of  the  electrostatic  force. 

The  difference  of  potential  between  two  points  is 


FIG.  9. — Equipotential 
surfaces. 


f 

Jo 


(12) 


where  ^  is  the  electrostatic  force  at  any  point  and  D  is  the  dis- 
tance between  the  points.     The  potential  gradient  at  any  point  is 

de  ,1Qv 

®  =  dr  = 

and  is  equal  to  the  electrostatic  force  at  the  point. 

In  practical  units  the  potential  gradient  is  expressed  in  volts 
per  centimeter  or  kilovolts  per  centimeter. 


10 


ELECTRICAL  ENGINEERING 


20.  Potential  and  Potential  Gradient  Near  a  Charged  Sphere. 

—  Determine  the  potential  and  potential  gradient  in  the  field 
surrounding  an  isolated  sphere  of  radius  R  charged  with  Q  units 
of  electricity  (Fig.  10). 

The  charge  Q  is  uniformly  distributed  over  the  surface  of  the 
sphere  and  may  be  considered  as  an  equal  charge  Q  concentrated 
at  the  center  (see  Art.  14). 

The  potential  at  a  point  P,  distant  r  cm.  from  the  center  of 
the  sphere  is 

Q 


e  =  — • 


The  potential  gradient  at  P  is 

de 


Q 


(14) 


and  is  equal  to  the  electrostatic  force  at  the  point. 


FIG.   10. — Potential  and  potential  gradient  near  a  charged  sphere. 

Curve  1,  Fig.  10,  shows  the  variation  of  potential  with  distance 
measured  from  the  center  of  the  sphere.  The  equation  of  this 
curve  is 

•-7-     •  - 

Across  the  sphere  the  potential  is  constant  and  is  equal  to  75; 
but  near  the  surface  it  falls  off  very  rapidly. 

The  potential  gradient  at  any  point  is  represented  by  the  slope 
of  the  tangent  to  the  potential  curve  at  that  point;  its  values  are 
plotted  in  curve  2.  The  equation  of  this  curve  is 

de  Q 


ELECTROSTATICS 


11 


21.  Potential  and  Potential  Gradient  between  Two  Charged 
Spheres. — Determine  the  potential  and  the  potential  gradient  at 
any  point  between  two  spheres  A  and  B  (Fig.  11)  of  radius  R  cm. 
if  the  distance  between  their  centers  is  D  cm.  and  they  have 
charges  of  Q  and  —  Q  respectively.  This  last  condition  means 
that  all  the  lines  of  dielectric  flux  which  leave  the  sphere  A  fall  on 
the  sphere  B. 


FIG.  11. — Potential  and  potential  gradient  between  two  charged  spheres. 

At  a  point  P  on  the  line  joining  the  centers  of  the  two  spheres 
and  at  a  distance  of  r  cm.  from  the  center  of  A  the  potential  due 

to  the  charge  Q  on  A  is  — ,  and  that  due  to  the  charge  —  Q  on  B 


s 


-Q 


;  the  actual  potential  at  P  is,  therefore, 


.Q  '     Q 

~  r       D  -r 


(15) 


Midway  between  the  spheres  the  potential  is 

Q  Q 


D/2 
At  the  surface  of  A  it  is 


D  -D/2 
Q 


=  0. 


(16) 


R       D  -  R 

The  potentials  at  all  points  between  A  and  B  are  plotted  in  curve 
1  (Fig.  11). 


12  ELECTRICAL  ENGINEERING 

The  potential  gradient  at  P  is 


_«.    _9_          /«  ,    jQ_\ 

r2        (Z)-r)2  \r2    h  (Z>  -  r)2/ 


Midway  between  the  spheres  the  potential  gradient  is 

/      0        I        0      \  8Q, 

V(D/2)2  t  (D/2)2/  "         D2 

which  is  its  minimum  value  and  at  the  surface  of  either  sphere  it 

(19) 


R2       (D  -  E)2 
its  maximum  value. 

The  potential  gradient  is  plotted  in  curve  2,  Fig.  11. 

It  has  been  assumed  that  the  charges  on  the  spheres  are  uni- 
formly distributed,  this  is  not  correct  as  may  be  seen  by  an  inspec- 
tion of  the  field  between  two  point  charges  in  Fig.  2.  The 
flux  and  the  charges  are  concentrated  on  the  sides  of  the  spheres 
which  are  facing  one  another  and  the  centers  of  the  charges  are 
drawn  together  through  a  very  short  distance.  Where  the  dis- 
tance between  the  spheres  is  large  compared  to  the  radius,  the 
error  is  small. 

22.  Capacity.  —  Conductors  of  different  sizes  and  shapes  have 
different  capacities  for  storing  electricity.  If  two  conductors  of 
different  capacities  are  charged  with  equal  quantities  of  electric- 
ity, they  will  be  raised  to  different  potentials,  the  one  of  small 
capacity  will  be  raised  to  a  high  potential  and  the  one  of  large 
capacity  to  a  low  potential. 

The  potential  to  which  a  body  is  raised  varies  directly  as  the 
quantity  of  electricity  stored  in  it  and  inversely  as  its  capacity 
and  may  be  expressed  by  the  formula 

E  =  |>  (20) 

where     Q  is  the  quantity  of  electricity,  or  charge, 
C  is  the  capacity  of  the  body, 
E  is  the  potential. 

The  capacity  therefore  is 

C  =  |  (21) 

and  is  equal  to  the  charge  divided  by  the  potential,  or  it  is  equal 
to  the  charge  per  unit  potential. 


ELECTROSTATICS  13 

A  body  has  unit  capacity  (electrostatic)  when  one  unit  of  elec- 
tricity is  required  to  raise  its  potential  by  unity. 

A  sphere  with  a  radius  of  1  cm.  has  unit  capacity,  because 
if  a  charge  of  one  unit  be  given  to  it  it  will  act  as  though  it 
were  concentrated  at  the  center  and  will  produce  at  the  surface 
unit  potential.  The  capacity  of  a  sphere  of  radius  R  cm.  is  R 
electrostatic  units. 

23.  Farad. — The  practical  unit  of  capacity  is  the  farad.  A 
conductor  has  a  capacity  of  one  farad  when  one  coulomb  of  elec- 
tricity is  required  to  raise  its  potential  by  one  volt. 

1  coulomb       3  X  109 


1  farad 


1  volt  }<joo 

9  X  1011  electrostatic  units.       (22) 


The  farad  is  a  very  large  unit  and  capacities  are  usually 
expressed  in  microfarads  or  millionths  of  a  farad. 

24.  Dielectric  Permeance. — The  foregoing  explanation  of 
capacity  is  apt  to  be  misleading.  A  conductor  has  capacity  only 
with  respect  to  surrounding  objects, 
since  the  electrostatic  energy  is  not 
stored  on  or  in  the  conductor  itself 
but  in  the  field  between  the  conductor 
and  surrounding  conductors. 

In  Fig.  12,  A  is  a  conductor  placed 
near    a    large    conducting    plane   B.  B~ 

Assume  a  potential  difference  E  to       FUIG-  12.— Permeance  of  the 

_____     path  between  a  conductor  and 

exist    between    A    and    B,    then    E    a  plane, 
measures  the  work  in  ergs  that  must 

be  done  in  carrying  a  unit  charge  from  B  to  A  against  the  elec- 
trostatic forces  in  the  field.  These  forces  produce  a  dielectric 
flux  passing  from  A  to  B. 

The  total  dielectric  flux  i/>  is  proportional  to  the  difference  of 
potential  E  and  to  the  permeance  of  the  path  (P. 

The  permeance  (P  is  directly  proportional  to  the  cross-sectional 
area  of  the  path  and  to  its  dielectric  constant  and  is  inversely 
proportional  to  the  length  of  the  path.  It  is  usually  difficult  to 
calculate  the  permeance  directly,  since  the  section  of  the  path 
varies  throughout  its  length.  In  such  cases  the  dielectric  flux 

t 

\I/  or  the  electric  charge  Q  =  j-  is  assumed  and  the  potential  dif- 
ference E  calculated  as  in  the  preceding  examples. 


14 


ELECTRICAL  ENGINEERING 


The  permeance  is  then 


""•I-T-4-*7' 


(23) 


and  is  equal  to  the  capacity  of  the  system  consisting  of  the  two 
conductors  and  the  dielectric  between  them  multiplied  by  4n-. 

25.  Condenser.  —  An  electrical  condenser  is  an  arrangement  of 
conductors  which  is  capable  of  storing  a  large  quantity  of  elec- 
tricity, or  which  has  a  large  capacity.  It  is  one  in  which  a  large 
flux  is  produced  when  a  given  potential  difference  is  applied  to 
its  terminals. 

The  capacity  of  a  condenser  is  measured  by  the  amount  of 
charge  necessary  to  raise  its  potential  by  unity,  and  is  therefore 
equal  to  the  positive  charge  on  it  at  any  time  divided  by  the 
potential  difference  between  its  terminals;  this  relation  is  ex- 
pressed by  the  equation 


26.  Parallel  Plate  Condenser.  —  Determine  the  capacity  of  a 
condenser  formed  of  two  parallel  plates  each  having  an  area  of 


FIG.  13. — Parallel  plate  condensers. 

A  sq.  cm.  separated  by  t  cm.  of  material  of  dielectric  constant  k 
Fig.  13  (a). 

If  a  difference  of  potential  E  is  applied  between  the  terminals 
a  dielectric  flux  ^  is  produced  passing  from  the  plate  of  high 

^ 
potential  to  the  plate  of  low  potential;  a  positive  charge  Q  =  j- 


theref ore  appears  on  one  plate  and  a  negative  charge  —  Q  =  — 


47T 

Q 


on  the  other;  the  density  of  the  charge  on  the  plates  is  q  =  -r 


ELECTROSTATICS  15 

units  per  square  centimeter,  and  the  dielectric  flux  density  be- 

4irQ       \b 
tween  the  plates  is  3D  =  faq  =  -j-  =  -^  lines  per  square  centi- 

meter. The  thickness  of  the  dielectric  is  assumed  to  be  so  small 
that  the  lines  of  flux  travel  directly  across  from  one  plate  to  the 
other. 

The  intensity  of  the  electrostatic  force  is 


(24) 
~  K  "  Ak 

The  difference  of  potential  between  the  plates  is  the  work 
done  in  carrying  a  unit  charge  from  one  plate  to  the  other  against 
the  force  ff;  it  is  therefore 


• 

Ak 
since  if  is  constant;  and  the  capacity  of  the  condenser  is 


C-Q 
E 


Q         Ak 

-  electrostatic  units.  (25) 


lirQt 
Ak 

The  capacity  therefore  varies  directly  with  the  area  of  the 
plates  and  with  the  dielectric  constant  of  the  material  separating 
them,  and  inversely  as  the  distance  between  them. 

When  the  plates  are  separated  by  air  the  capacity  is 

' 


In  order  to  increase  the  capacity  of  such  a  condenser  a  large 
number  of  plates  are  used  joined  in  multiple  and  separated  by 
very  thin  sheets  of  dielectric  as  shown  in  Fig.  136. 

27.  Capacity  of  Concentric  Cylinders.  —  Determine  the  capac- 
ity of  a  condenser  formed  of  two  concentric  cylinders,  Fig.  14, 
of  radii  Ri  and  R2  cm. 

If  a  charge  of  q  units  per  centimeter  length  is  given  to  the 
inner  cylinder,  lines  of  dielectric  flux  pass  out  radially  and  pro- 
duce, at  a  distance  r  cm.  from  the  axis  of  the  cylinder,  a  flux 
density 


2<7  ,. 

3D  =  -j—  =  —  lines  per  square  centimeter. 


16  ELECTRICAL  ENGINEERING 

If  the  dielectric  constant  of  the  material  is  k,  the  electrostatic 
force  at  the  point  is 

D 


and  the  difference  of  potential  between  the  two  cylinders  is 


The  capacity  per  centimeter  length  of  the  condenser  is 

C  =  ^  =  ^  -  £-=-  =  -  ^-=-  electrostatic  units.       (28) 


This  is  the  case  of  a  single-conductor  cable  with  a  lead  sheath, 
and  the  capacity  is  usually  expressed  in  farads  per  mile;  it  is 

r         _k  __       2.54  X  12  X  5,280 

2iog|  9X10U  :v 

k  2.54  X  12  X  5,280 

/\ 


2  X  2.303  log,  % 

Rl 

=  3.82  X  —  ^-5-  X  10-6  farads  per  mile.  (29) 

,         ri2 
logic  «- 
KI 

28.  Potential  and  Potential  Gradient  between  Concentric 
Cylinders.  —  Assume  that  the  potential  of  the  inner  cylinder  in 
Fig.  14  is  E  and  that  of  the  outer  one  is  zero. 

The  charge  per  centimeter  length  is 

q  =  CE  =  —  ^-=-  by  equation  (28). 

2  log  f2 
Hi 

The  potential  at  radius  r  is 

E     dr 


(* 

J< 


,  2 

««        log  "7 
[logr]     -£  --  £•  (30) 


ELECTROSTATICS 


17 


The  potential  gradient  at  radius  r  is 

=  £??  =  ?2  =     _A 

dr       AT 


(31) 


it  has  its  maximum  value  at  the  surface  of  the  inner  cylinder. 

E 

{/max,   —  D 


(32) 


Curve  1,  Fig.  14,  shows  the  variation  of  the  gradient  from  one 
conductor  to  the  other. 

The  area  under  the  curve  represents  the  difference  of  potential, 
E,  between  the  conductors. 


4.        Curve  1,0  — 


Area  «=  Difference  of  Potential  between  Cylinder! 
^-  Potential  =0 


-Potential  =  E 


k  —  Dielectric  Constant 


FIG.  14.  —  Concentric  cylinders. 

The  ends  of  the  outer  cylinder  are  shown  turned  out  at  a 
radius  not  less  than  the  distance  between  the  conductors.  This 
prevents  concentration  of  flux  at  the  ends  which  might  cause  a 
breakdown  if  the  difference  of  potential  were  very  great. 

29.  Capacity  of  Parallel  Conductors.  —  Determine  the  capacity 
of  two  parallel  cylindrical  conductors  A  and  B,  Fig.  15(6)  of 
radius  R  cm.  suspended  in  air  at  a  distance  of  D  cm.  between 
centers. 

When  a  difference  of  potential  E  is  applied  between  them  a 
dielectric  flux  \j/  passes  across  from  A  to  B  and  a  positive  charge 

t 

Q  =  -  -  appears  on  A  and  an  equal  negative  charge  —  Q  =  — 

~r7T 


If  q  is  the  charge  per  centimeter  length  on  A,  the  electrostatic 
force  at  any  point  Q,  on  the  line  joining  the  centers,  distant  r  cm. 
from  A  and  D  —  r  cm.  from  B  is  the  resultant  of  the  forces  exerted 
by  the  charges  on  A  and  B  acting  independently. 


18 


ELECTRICAL  ENGINEERING 


From  each  centimeter  length  of  A,  4^-q  lines  of  dielectric  flux 
extend  out  normally  and  produce  at  the  point  Q  a  flux  density 

47T#  20  .-. 

£>A  =  2^:  =:  -^  lmes  Per  square  centimeter. 
The  electrostatic  force  at  the  point  Q  is 


and  acts  from  A  to  B. 


=  ~  dynes 


FIG.  15. — Electrostatic  field  between  two  parallel  cylindrical  conductors. 

The  charge  —  q  on  B  produces  at  Q  a  force 


in  the  same  direction  as  &A. 
The  resultant  force  at  Q  is 


2? 


dynes. 


(33) 


The  work  done  in  moving  a  unit  positive  charge  from  B  to  A  is 
equal  to  the  difference  of  potential  between  the  two  conductors; 
it  is 

-  R 


" 


0r,D-jR 
29[log  -^-    - 


, 

=  4q  log 


(34) 


ELECTROSTATICS  19 

The  capacity  of  the  two  conductors  per  centimeter  length  is 
C  =  ^  =  ^ n  =  — — ~ =5  electrostatic  units.    (35) 

Mi          .     ,       Lf  —  K         ,  ,        D  —  ft 
±q  log  — ^—        4  log  — £— 

The  capacity  per  mile  of  a  line  consisting  of  two  conductors  is 
c  =  2.54  X  12  X  5,280 


4  X  2.303  log™  D  p  R  X  9  X1011 

£1 

19.4  X  10~9  .       ,  ,0fl, 
=r =•  farads.                                        (36) 

i  Iy  —  it 

log* -a-     ^ 


It  is  sometimes  useful  to  separate  the  capacity  of  the  line  into 
the  capacities  of  the  two  conductors  forming  it.  The  potential 
of  the  point  0  midway  between  the  conductors  is  zero  and  the 
potential  of  conductor  A  is 


. 

and,  therefore,  the  capacity  of  A  per  centimeter  length  between 
the  conductor  and  the  neutral  or  point  of  zero  potential  is* 

CA  =  ~  =  -      —±j?  -  5  =  —  —  =r  --  5  electrostatic  units  (37) 
Hi  A        n    i       Lf  —  K        n  .       L)  —  K 
2q  log  —  ^—        2  log  —  -£- 

and  the  capacity  of  each  conductor  in  farads  per  mile  is 

_  38.8  X  10-»  • 

D-R 

logM  -^ 

30.  Potential  and  Potential  Gradient  between  Parallel  Con- 
ductors. —  The  potential  at  the  point  Q,  Fig.  15(6),  is  the  work 
done  in  moving  a  unit  positive  charge  from  the  neutral  point  0  to 
the  point  Q;  it  is 


e  = 


20  ELECTRICAL  ENGINEERING 

,         ,  TjJ 

q  =  CE  =  -      —  yr  ---  -  from  equation  (35), 


and,  therefore, 

1  r\cr  _ 

•  -  2  *  ^  ~  *"  '-  (39) 


This  equation  is  plotted  in  curve  1,  Fig.  15  (a). 

TjJ  77f 

The  potential  of  conductor  A  is  —  and  that  of  B  is  —  —  • 

2  2 

The  potential  gradient,  at  any  point  Q,  which  is  equal  to  the 
intensity  of  the  electrostatic  field  at  the  point,  is  given  by  the 
equation 


or  expressed  in  terms  of  the  potential  of  the  conductor  A  it  is 

*-    *' 


,      D-R(r(D- 

log-?r 

For  small  values  of  r  and  72  this  may  be  written 

EA 
9  =  ~      —J^' 


If  E  and  ^A  are  expressed  in  volts  the  potential  gradient  0  is 
expressed  in  volts  per  centimeter. 

Equation  40  is  plotted  in  curve  2,  Fig.  15  (a). 

If  E  is  taken  as  200,000  volts,  D  as  200  cm.  and  R  as  1  cm., 
the  maximum  gradient,  which  occurs  at  the  surface  of  the  con- 
ductor is 

'  '  (43) 


=  19'000  volts  per  centimeter  and  the 


ELECTROSTATICS  21 

minimum  gradient,  which  occurs  at  the  point  midway  between  the 
conductors,  is 

D 


D  -  R 


D      D 

2  X   2 


Dlog 


-  R 


R 


v  ,   1        _  onn      1A 
X  -^  =  3,800  volts  per  centimeter. 
log  199        5U 

If  the  two  conductors  had  been  suspended  in  a  medium  of 
dielectric  constant  k  and  the  same  difference  of  potential  applied 
between  them  the  capacity  and  the  charge  per  centimeter  would 
have  been  increased  in  the  ratio  k  but  the  potential  and  potential 
gradient  would  remain  as  before. 

31.  Dielectric  Field  and  Equipotential  Surfaces  between  Par- 
allel Conductors.  —  The  field  between  two  parallel  conductors  is 
shown  in  Fig.  15(d).  It  is  obtained  by  superposing  the  fields  due 
to  the  conductors  A  and  B  separately.  At  the  point  P  in  Fig. 

15(6)  the  two  forces  acting  are  /i  =  —  due  to  the   charge  on  A 

and  /2  =  —  due  to  the  charge  on  B.     These  two  forces  may  be 

combined  vectorially  to  give  the  resultant  force  3>,  which  is 
tangent  to  the  line  of  force  passing  through  P.  The  lines  of 
force  are  excentric  circles  cutting  the  conductors  normally  and 
passing  through  points  slightly  displaced  from  their  centers. 

A  number  of  equipotential  surfaces  are  also  shown  in  Fig. 
15  (d)',  they  are  excentric  cylinders  surrounding  the  conductors 
with  their  centers  on  the  line  joining  the  conductor  centers. 

The  equations  of  these  surfaces  may  be  found  as  follows: 

In  Fig.  15(c)  the  point  P  has  coordinates  r*i  and  r2  referred  to 
the  centers  of  the  two  conductors,  and  coordinates  x  and  y  re- 
ferred to  the  rectangular  axes.  It  is  required  to  determine  the 
equation  of  th£  equipotential  surface  passing  through  P. 

The  potential  at  P  is 


e  = 


ri  r2 

2CE  log  (45) 


22  ELECTRICAL  ENGINEERING 

and  ,  .      r2          e 


or 

—  =  6  2CE  —  fc  _  a  constant  for  the  surface. 
?*i 

Expressing  r\  and  r2  in  terms  of  a:  and  y, 

r2  =  ATI  or  r22  =  /b2ri2 
(D  -  xY  +  y2  =  k2  (x2  +  y2) 

or,  simplifying, 

(A;2  -  1)  x2  +  2Dx  +  (k2  -  1)  y2  =  D2 

2         2D      .  ,     2  =       D2 

X  +  fc2  -  1  x  +  y    "  fc2  -  1 


-!      '    (fc2-l)2  '  kz  -  1    '    (fc2-!)2      (fc2-!)2 


This  is  the  equation  of  a  circle  with  its  center  at  the  point 

and  of  radius       2  _       '  (47) 


Thus  the  equipotential  surfaces  are  cylinders  (the  traces  in  the 
xy  plane  are  circles)  and  may  be  determined  by  assuming  any 

e 

potential  e  and  substituting  the  value  k  =  €2CE  in  the  equations 
above. 

32.  Capacity  of  a  Single  Wire  to  Earth.  —  Determine  the  capac- 
ity of  a  single  wire  of  radius  R  cm.,  suspended  at  a  height  H  cm. 
above  the  earth.  If  the  wire  is  raised  to  a  potential  E  above 
the  earth  potential,  lines  of  dielectric  flux  will  pass  from  the  wire 
to  earth  as  shown  in  Fig.  16.  If  ^  is  the  flux  from  each  centi- 

& 
meter  length  of  A,  then  q  =  j-  is  the  charge  on  each  centimeter 

of  A,  and  a  corresponding  negative  charge  —  q  appears  on  the 
earth,  but  it  is  not  evenly  distributed  being  most  concentrated 
directly  beneath  the  wire. 

The  flux  passing  from  the  wire  A  to  the  earth  would  be  un- 
changed if  the  charge  on  the  earth  were  collected  on  a  second 
wire  B  placed  at  a  distance  H  cm.  below  the  earth.  Its  potential 
would  be  —  E. 


ELECTROSTATICS 


23 


The  difference  of  potential  between  A  and  B  is  2E  and  from 
equation  (34) 


2E  =  4q  log 


2H  -  R 


thus  the  capacity  of  the  single  wire  with  respect  to  the  earth  per 
centimeter  length  is 

a  1 


r  -  q 
C  ~ 


n   .      2H  -  R    "      .       2H  -  R 
2glog     ~^-          2  log 


e.s.   units,      (48) 


R  R 

and  the  capacity  in  farads  per  mile  is 

38.8  X  10~9 


rrteth-n  ' «'. 'isi'iii"^.')  'I/1 


•v 


FIG.  16. — Capacity  of  a  wire 
to  earth. 


FIG.  17. — Capacity  of  a  sphere 
to  earth. 


33.  Capacity  of  a  Sphere  to  Earth.  —  In  Fig.  17,  A  is  a  sphere 
of  radius  R  cm.  at  a  height  H  cm.  above  the  earth.  If  A  is 
raised  to  a  potential  E  above  earth  potential,  a  dielectric  flux 
\I/  is  produced  passing  from  the  sphere  to  the  earth  as  shown 

\L 

and  a  positive  charge  Q  =  j-  appears  on  A  and  an  equal  negative 

4?r 

charge  on  the  earth. 

Without  changing  the  distribution  of  flux  in  any  way  the  nega- 
tive charge  on  the  earth  may  be  assumed  to  be  collected  on  a 
sphere  B  similar  to  A  but  placed  at  a  distance  H  cm.  below  the 
surface.  The  distribution  of  potential  between  two  such  spheres 
was  worked  out  in  Art.  21.  The  potential  midway  between 
them  is  zero,  the  potential  at  the  surface  of  A  is 


E  =      "" 


-  R  by  e(luation 


24  ELECTRICAL  ENGINEERING 

and  therefore  the  capacity  of  the  sphere  is 
„  _  Q  _  Q  1 


E      Q 
R 


Q 


1 


electrostatic  units.  (50) 


2H  -  R       R      2H   -  R 
When  H  is  very  large  compared  to  R  the  capacity  is 

as  in  Art.  22. 


34.  Condensers  in  Multiple.  —  If  a  number  of  condensers  of 
capacities  Ci,  C2  and  Cz  are  connected  in  multiple,  as  shown  in 
Fig.  18,  and  a  difference  of  potential  E  is  applied  to  the  terminals, 
each  condenser  receives  a  charge  proportional  to  its  capacity, 

Qi  =  CiE, 
Qz  =  C-2.E, 

Qs  = 


and  the  total  charge  on  the  system  is 

Q  =  Qi  +  Q*  +  Q3  =  E  (Ci  +  C2  +  C3). 


Foffe 


Fio.  18. — Capacities  in  multiple.  FIG.  19. — Capacities  in  series. 

The  capacity  of  the  system  is 


(51) 


and  therefore  the  capacity  of  a  number  of  condensers  connected 
in  multiple  is  equal  to  the  sum  of  their  separate  capacities. 

35.  Condensers  in  Series. — When  a  number  of  condensers  of 
capacities  Ci,  Cz  and  Cs  are  connected  in  series,  as  in  Fig.  19,  and 
a  difference  of  potential  E  is  applied  to  the  terminals  of  the  sys- 
tem, a  charge  Q  appears  on  each  condenser  and  the  potential  E 


ELECTROSTATIC^  25 

divides  up  among  the  condensers  in  inverse  proportion  to  their 
capacities. 

The  drop  of  potential  across  condenser  (1)  is 

Q 

that  across  (2)  is 

Et,o_, 

and  that  across  (3)  is 

Es  =  Q_. 

but 

771  T7T  I         T7T  I         T7T 

&      =   til  -\-  &2  -\-  J&s 
1 


and,  therefore,  the  capacity  of  the  system  consisting  of  three  con- 
densers in  series  is 

r  _  Q  __  L  . 

J   -  +  -  +  - 

Ci          L>2          ^8 

When  two  equal  condensers  of  capacity  C  are  connected  in 
series,  their  combined  capacity  is 

(53) 


i 

C  +  C 

and  is  equal  to  one-half  of  the  capacity  of  either  condenser  alone. 
38.  Energy  Stored  in  a  Condenser.  —  When  a  condenser  is 
being  charged,  work  is  done  in  rais- 
ing the  charge  through  the  difference 
of  potential  between  the  terminals, 
and  this  amount  of  energy  is  stored 
in  the  electrostatic  field  of  the  con- 


Fio.  20. 
denser. 

In  Fig.  20,  PN  is  a  condenser  of  capacity  C  formed  of  two 
parallel  plates  separated  by  t  cm.  of  a  dielectric  of  constant  k. 
When  a  potential  difference  e  is  produced  between  the  plates  by 
the  generator  G,  electricity  flows  from  N  to  P  until  the  charge 
on  P  is 


26  ELECTRICAL  ENGINEERING 

if  the  potential  e  is  increased  by  de  the  charge  q  is  increased  by 
dq  =  C  de  and  the  work  done  in  raising  the  charge  dq  through 
the  difference  of  potential  e  is 

dw  =  e  dq  =  Ce  de. 

The  total  work  done  in  charging  the  condenser  with  a  quantity 
of  electricity  Q,  or  to  a  difference  of  potential  E,  is 

W  =  I  dw  =  f  Ce  de 
J  Jo 

•~*v  -c[lt 

=  C  ~  ergs.  (54) 

Thus  the  work  done  in  charging  a  condenser,  or  the  energy 
stored  in  the  electrostatic  field  of  the  condenser,  is  equal  to  one- 
half  of  the  capacity  multiplied  by  the  square  of  the  difference  of 
potential  between  the  terminals. 

Equation  (54)  may  be  expressed  in  two  other  forms  by  substi- 
tuting for  E  its  value  -^', 


W  =     QE,  (55) 

W  =i|2-         .   .  '       (56) 

If  the  area  of  the  plate  P  is  A  sq.  cm.,  then  the  dielectric  flux 
density  between  the  plates  is 


3D  =  —  -p  lines  per  square  centimeter, 

and  the  electrostatic  force  is 

^  =  T  dynes. 

The  potential  difference  between  the  plates  is 


and  the  capacity  of  the  condenser  by  equation  (25)  is 

Ak 
L  "47? 


ELECTROSTATICS  27 

Substituting  these  values  for  E  and  C  in  equation  (54)  gives 
a  fourth  expression  for  the  energy  stored  in  the  field,  namely 

w  =  \ x  55 x  (r 

*£  rrvrt  \  Iv  . 

-  AI.X  s=t  ergs.  (57) 


Since  the  volume  of  the  field  is  At  c.c.  and  the  flux  density  is  uni- 
form, the  energy  stored  per  cubic  centimeter  of  the  field  is 

3D2 


W  = 
or 

$zk 
w  =  -g^-  ergs.  (59) 

Thus,  the  energy  stored  per  cubic  centimeter  in  an  electro- 
static field  is  equal  to  the  square  of  the  dielectric  flux  density 

multiplied  by  ^-y,  or  is  equal  to  the  square  of  the  intensity  of 


k 
the  electrostatic  force  multiplied  by  TT-* 

O7T 

From  equation  (54)  a  very  useful  definition  of  capacity  may  be 
obtained, 

C  =  ^  (60) 

or  the  capacity  of  a  condenser  is  equal  to  twice  the  energy  stored 
in  its  field  divided  by  the  square  of  the  difference  of  potential 
across  its  terminals,  or  the  capacity  is  equal  to  twice  the  energy 
stored  when  the  difference  of  potential  is  unity. 

37.  Stresses  in  an  Electrostatic  Field.  —  The  energy  stored  in 
an  electrostatic  field  is 

E2 
W  =  C~  ergs; 

and  the  energy  stored  per  cubic  centimeter  is 

£>2 
*,=  —  ergs. 

These  two  equations  represent  the  potential  energy  of  the  field. 
Stresses  exist  throughout  the  field  tending  to  reduce  the  potential 
energy  to  a  minimum;  first,  there  is  a  tension  along  the  lines  of 
induction  tending  to  shorten  them  and  to  draw  the  bounding 


28  ELECTRICAL  ENGINEERING 

surfaces  of  the  field  together  and  so  reduce  the  volume  to  zero; 
second,  there  is  a  pressure  at  right  angles  to  the  lines  tending 
^.^ mim mm^  to  spread  them  apart  and  so  reduce  the 
"£"  density  in  the  field.  Since  the  system  is  in 
equilibrium  these  two  stresses  are  of  equal 
magnitude. 

To  obtain  an  expression  for  the  stress  per 
FIG  21  square  centimeter  on  the  bounding  surfaces, 

consider    the    parallel    plate    condenser    in 
Fig.  21.     The  energy  stored  in  the  field  is  by  equation  (57) 

T»2 

W  =  At  X  A 


i: 


If  a  force  of  P  dynes  is  applied  to  one  of  the  plates  and  the  dis- 
tance between  the  plates  is  increased  by  amount  dt,  the  work 
done  is  P  dt  ergs.  The  charges  on  the  plates  are  assumed  to  re- 
main constant  and  therefore  the  flux  density  remains  constant, 
but  the  volume  of  the  field  is  increased  by  the  amount  A  dt  c.c., 

3D2 
and  the  energy  stored  in  it  is  increased  by  A  dt  X  TT-T,  but  the 

increase  in  the  stored  energy  is  equal  to  the  work  done  by  the 
force  P  and,  therefore, 

T)2 

Pdt  =  Adt  X^ 
and 

P  =  A  X  J^  dynes.  (61) 


This  is  the  pull  exerted  by  the  field  on  each  plate  of  the  condenser 
tending  to  draw  them  together. 
The  pull  per  square  centimeter  is 

(62) 


thus,  the  pull  per  square  centimeter  on  any  charged  surface  is 
equal  to  the  square  of  the  induction  density  at  the  point  divided 
by  Sirk. 

This  is  the  value  of  the  tension  along  the  lines  of  force  tending 
to  shorten  them,  and  also  the  value  of  the  pressure  at  right  angles 
to  the  lines  tending  to  spread  them  apart. 

38.  Force  Exerted  on  a  Dielectric  by  an  Electrostatic  Field. 
—  Fig.  22  shows  two  metallic  plates  of  area  A  sq.  cm.  separated 


ELECTROSTATICS  29 

by  a  dielectric  of  constant  k  and  thickness  t  cm.  Determine 
the  pull  required  to  remove  the  dielectric  from  the  condenser 
against  the  opposing  force  due  to  the  field. 

(a)  Assume,  first,  that  the  condenser  has  a  charge  Q  and  is  not 
connected  to  any  source  of  potential. 

The  energy  stored  in  the  condenser  is 

4   v/ 1  ^  xi /L/  J\. K 

when  the  dielectric  is  removed  the  stored  energy  is 
Wz  =  2  Cl  =  2  ~A   : 


where  C\  is  the  capacity  before  and  C2  is  the  capacity  after  the 
dielectric  is  removed. 


Dielectric  I 

Constant    k  !"'*''* 


-Q 


3 

Dielectric 
Constant   & 

I 
h 

1 

1 

,                J 

d               1 

(a)  FIG.  22.  (6) 

The  change  of  stored  energy  is 

W        W     -  2r^2  /i  _  IN\ 
A     \     ~  k)' 

and  this  is  equal  to  the  average  pull  P  multiplied  by  th*»  distance 
through  which  it  is  exerted  =  Pd. 
Therefore,  the  average  pull  is 


Z>  *  ''I  *Jiti/\g     /^  JL  v      _.  -jj,.         f)O/     \  ^AQ\ 


but  Q  =  Ci^Ji  =     .    . 1 ,   where  Ei  is  the  initial  difference  of 

potential  between  the  plates;  and,  therefore, 

9-7T/  A  2fr2  T^2/  1\         ATc  (1?  1^7^2 

D£iT\(i  xj.    Iv  JCfl     /4  X\  ^1/v  v./v  J.>ZI/i         .  /r>^\ 

^"l1  ~  jfc)  =  ~         STT^          dyneS<        (6 4) 


Ad 
The  final  potential  difference  between  the  plates  is 


C 


30  ELECTRICAL  ENGINEERING 

(by  Assume,  second  that  the  difference  of  potential  EI  is 
maintained  constant  by  a  generator,  Fig.  22(6),  and  find  the 
pull  required  to  remove  the  dielectric. 

Initial  stored  energy  is 

irs*Ci*£.  £»*>!; 

the  final  stored  energy  is 

C2 ~2T    =  4^  ~2~" 
The  loss  of  stored  energy  is  / 


This  amount  of  energy  is  given  back  to  the  generator  supply- 
ing the  charge  and  as  before  the  average  pull  required  is 


_  _  . 

d  ' 


The  ratio  of  the  pulls  required  in  the  two  cases  (a)  and  (6)  is 

Ak  (k  -  1)  ffi2 
P_  Sirtd 

P1  ==   A  (k  -  1)  ES  '' 

Sirtd 

39.  Effects  of  Introducing  Dielectrics  of  Various  Specific  In- 
ductive Capacities  into  a  Uniform  Field.  —  1.  Fig.  23  shows  a 
parallel  plate  condenser  with  air  as  dielectric. 

t  =  distance  between  plates  in  centimeters. 
A  =  area  of  each  plate  in  square  centimeters. 
E  —  difference  of  potential  between  plates. 
^  =  dielectric  flux. 

^ 
3)  =  -7  =  dielectric  flux  density. 

fL 

$  =  3D  =  electrostatic  force  in  the  field. 

Since  the  force  is  constant  throughout  the  field,  therefore, 

E  =  $t 
and  the  electrostatic  force  or  the  stress  in  the  air  is 


If  E  is  expressed  in  volts,  the  potential  gradient  or  the  stress 
may  be  expressed  as 

E 

g  =  —  volts  per  centimeter. 


ELECTROSTATICS 


31 


Assuming  that  E  —  25,000  volts  and  t  =  1  cm.  the  gradient  or 

E      25  000 

stress  is  g  =  y  =  — ^ —  =  25,000  volts  per  centimeter.     Air  has 

L  -L 

a  dielectric  strength  of  31,000  volts  per  centimeter  and  it  will 
therefore  not  break  down  in  this  case. 

2.  In  Fig.  24  a  sheet  of  glass  of  thickness  0.3£  and  dielectric 
constant  k  =  6  is  introduced  into  the  field  as  shown  and  the 
difference  of  potential  is  the  same  as  before. 


K=6, 


ft 


1|LJ| 

Or  21 


FIG.  23. 


FIG.  24. 


FIG.  25. 


The  dielectric  flux  density  3D  is  constant  throughout  the  field; 
the  electrostatic  force  in  the  air  is 


the  drop  of  potential  across  the  air  portion  of  the  field  is 

El  =  Q.7t  X  5A  =  0.7*  X  3D; 
the  electrostatic  force  in  the  glass  is 

-  5       j? 
=  k  ==  6' 

The  drop  of  potential  across  the  glass  is 


=  0.3*  X 


X  if  =  0.05*  X 
o 


the  difference  of  potential  between  the  plates  is 

E  =  EI  ~\~  EZ 

=  0.7J  X  3D  +  0.05*  X  £> 


and  the  dielectric  flux  density  is 


0.75Z 


thus  the  stress  in  the  air  is 


and  is  greater  than  it  was  in  the  first  case. 

Again,  assuming  E  =  25,000  volts  and  t  =  1  cm.,  the  gradient 

in  the  air  is  g  =  |  -  =  ^  X  25,000  =  33,000  volts  and  the  air 

616  ,t 

will  break  down. 


32  ELECTRICAL  ENGINEERING 

In  this  case  by  introducing  a  dielectric  of  high  specific  inductive 
capacity  and  much  greater  dielectric  strength  than  the  air,  the 
air  is  made  to  break  down.  It  now  becomes  a  conductor  and 
the  full  stress  of  25,000  volts  comes  on  the  glass.  The  potential 

25  000 
gradient  in  the  glass  becomes  g  —      '  Q     =  83,000  volts   per 

U.o 

centimeter. 

3.  Fig.  25  shows  the  same  pair  of  plates  with  three  sheets  of 
dielectric  introduced  between  them,  of  thickness  ti,  fa  and  ts  and 
dielectric  constants  fci,  &2  and  &3  respectively. 

3D  =  dielectric  flux,  which  is  constant  throughout  the  field. 

3D 
^i  =  •=-  =  stress  in  layer  (1), 

/C] 

«F2  =  --  =  stress  in  layer  (2), 
K% 

3D 

$3  =  Y~  =  stress  in  layer  (3), 

#3 

3D 

EI  =  $iti  —  j-  ti  —  drop  of  potential  across  (1), 
KI 

3D 

Ez  =   g^2  =  —  fa  =  drop  of  potential  across  (2), 
KZ 

3D 
E3  =   3^3  =  =- 13  =  drop  of  potential  across  (3); 

&3 

the  difference  of  potential  between  the  plates  is 
E  =  Ei  +  #2  +  #3 


'      If         '      1? 
/x/2  fv 

and  the  dielectric  density  is 

E' 


h  +  b  +  !i 

and  the  total  flux  is 

— ? -A=E<5> 


where 


l      ,         2     , 
fci  fa  fc3 


<P  =       =  7  -     "  T  ecluation 
•"       A1  4.  :?.  4.  ?L 


is  the  permeance  of  the  path  between  the  plates. 


ELECTROSTATICS 
The  drop  of  potential  across  (1)  is 


33 


KI  /  ti  .    t%  .    ts\ 

—  (  T~  ~r  —  n I 

r-  \  £*  I/*  Is*     / 

I/I   \M'l  A/2  A'3/ 


and  the  potential  gradient  is 
_  E,  _ 

01  ="    : 


E 


=1  (£  + ,-  +  ^ 


(66) 


(67) 


The  potential  gradients  in  (2)  and  (3)  may  be  expressed  by 
similar  equations. 

From  the  last  two  examples  it  is  seen  that  a  uniform  field  can 
be  made  non-uniform  by  introducing  dielectrics  of  various  specific 
inductive  capacities,  and  that  the  stresses  in  the  various  dielec- 
trics vary  inversely  as  their  specific  inductive  capacities. 

4.  In  Fig.  26  a  block  of  dielectric  constant  k  is  introduced  into 
the  uniform  field  as  shown. 

The  stress  in  the  air  is 

-  E 
**  ="  1' 

and  is  the  same  as  before  the  dielectric  was  introduced. 
The  stress  in  the  dielectric  is  also 

=  - 

D   ~         A  ~     t, 

but  the  flux  density  in  the  dielectric  is 

ET 

£>  =  $Dk  =  —k  =  5Ak. 


FIG.  26. 


t 


FIG.  27. 


I 

„*•- 
--* 

a 

d 
f 

FIG.  28. 


The  permeance  of  the  path  is  increased  and  an  increased  flux 
is  produced  passing  between  the  plates,  but  since  the  increased 
flux  is  all  confined  to  the  dielectric  there  is  no  increase  in  the  stress 
in  the  air. 

5.  In  Fig.  27,  the  same  block  of  dielectric  is  shown  with  a 
groove  cut  in  it.  The  air  in  the  groove  is  very  highly  stressed 


34 


ELECTRICAL  ENGINEERING 


and  as  the  difference  of  potential  between  the  plates  is  increased, 
this  air  will  be  the  first  to  break  down. 

6.  Fig.  28  is  another  example  of  the  same^ thing.     The  stress  on 
the  air  in  the  pockets  a  and  b  is  high.     As  the  difference  of  poten- 
tial is  increased  this  air  will  break  down  and  become  conducting 
and  there  will  be  a  discharge  over  the  surface  acb  at  a  potential 
difference  much  lower  than  that  required  to  break  over  the  shorter 
surface  df,  where  the  stress  is  uniform. 

7.  In  Fig.  29,  A  is  a  small  piece  of  material  with  a  high  dielectric 
constant  such  as  a  drop  of  oil.     It  offers  a  local  path  of  high 


FIG.  29. 


FIG.  30. 


FIG.  31. 


FIG.  32. 


permeability,  and  a  greater  flux  density  is  produced  in  it  than  in 
other  parts  of  the  field,  and  therefore  the  stress  in  the  air  at  its 
surface  will  be  greater  than  the  average  stress  throughout  the 
field. 

8.  In  Fig.  30  the  drop  of  oil  is  replaced  by  a  knob  on  the  con- 
ductor forming  one  boundary  of  the  field.     Since  the  dielectric 
constant  of  the  conducting  knob  is  infinity  the  stress  in  the  air  at 
its  surface  will  be  greater  than  in  the  case  of  the  oil. 

9.  Fig.  31  shows  the  field  at  the  edge  of  the  condenser  in  case 
(1).     In  this  region  the  dielectric  flux  density  is  not  uniform,  and 
just  at  the  edges  of  the  plates  it  is  greater  than  in  the  main  body 
of  the  field.     The  electrostatic  stress  is  also  greater  than  the 
average  value. 

This  condition  may  be  corrected  by  turning  out  the  edges  of 
the  plates  or  electrodes  at  a  radius  not  less  than  the  distance 
between  them  as  shown  in  Fig.  32. 

40.  Effect  of  Introducing  Conductors  into  Electrostatic  Fields. 
— An  insulated  conductor  may  be  introduced  into  an  electrostatic 
field  without  changing  the  flux  distribution  if  it  is  placed  entirely 
on  an  equipotential  surface.  Fig.  33(6)  shows  such  a  case. 
The  conductor  takes  the  potential  of  the  surface  on  which  it  lies. 

In  Fig.  33  (c)  the  conductor  does  not  lie  along  one  of  the  original 
equipotential  surfaces  and  the  distribution  of  flux  is  changed  as 


ELECTROSTATICS 


35 


shown.  The  conductor  is  an  equipotential  surface  and  the  lines 
of  force  enter  it  normally.  Its  potential  is  fixed  by  the  relative 
capacities  between  it  and  the  two  plates  forming  the  boundaries 
of  the  field.  If  it  is  placed  symmetrically  between  the  plates 
its  potential  is  the  mean  of  the  potentials  of  the  plates. 

In  Fig.  15  which  represents  the  field  between  two   parallel 
cylindrical  conductors  an  insulated  conducting  cylinder  may  be 


0  Volts 


(a) 


0  Volts 


placed  on  any  one  of  the  equipotential  surfaces  without  changing 
the  field;  and  if  this  cylinder  is  then  connected  to  a  source  of 
potential  of  the  same  value,  the  original  conductor  may  be  re- 
moved without  changing  the  field  external  to  the  cylinder. 

41.  Graded  Insulation  for  Cables. — Fig.  34 (a)  shows  a  single- 
conductor  cable  insulated  with  a  dielectric  of  constant  ki  =  6; 


Maximum  Gradient 


FH«H 

O)  (6) 

FIG.  34. — Graded  insulation  for  cables. 

the  radius  of  the  conductor  is  R\  =  0.4  cm.  and  the  inside  radius 
of  the  sheath  is  R*  =  2.4  cm.  If  the  sheath  is  grounded  and  the 
conductor  is  raised  to  a  potential  E,  the  potential  at  any  point 
in  the  dielectric  at  a  distance  r  cm.  from  the  center  of  the  con- 
ductor is 


e  =  E 


,     R* 
log- 


by  equation  (30) 


36  ELECTRICAL  ENGINEERING 

and  the  potential  gradient  or  stress  at  the  point  is7 

g  =  -T-  =  -     J—~-  volts  per  centimeter  by  equation  (31). 
dr         ,      rt4 


The  gradient  is  plotted  in  the  figure  and  the  hatched  area  under 
the  curve  represents  the  difference  of  potential  E. 

The  maximum  gradient  or  stress  occurs  at  the  surface  of  the 
conductor  where  r  =  R\\  it  is 


i  log 


(68) 


If  the  maximum  allowable  stress  is  assumed  to  be  100,000  volts 
per  centimeter,  then  the  maximum  difference  of  potential  at 
which  the  cable  can  be  operated  safely  is 


=  100,000  X  0.4  log  ~    =  72,500  volts.  (69) 

In  this  case  the  stress  in  the  outer  layers  of  the  dielectric  is  far 
below  the  maximum  allowable  stress  and  the  material  is  not 
used  to  the  best  advantage. 

The  capacity  of  this  cable  per  centimeter  length,  equation  (28) 
is 

C  =  -     ^-5-  =  -     —  ^-7  =  1.585  electrostatic  units. 
21og|       21og|| 

Fig.  34(6)  shows  the  same  cable  insulated  with  three  layers  of 
materials  of  dielectric  constants  ki  =  6,  A;2  =  4  and  kz  =  2; 
the  outside  radii  of  the  three  layers  are  R2  =  0.6,  Rs  =  1.2  and 
fl4  =  2.4  cm. 

If  q  is  the  charge  per  centimeter  length,  the  dielectric  flux  den- 
sity at  radius  r  is 

3D  =  —  lines  per  square  centimeter 
and  the  electrostatic  force  or  stress  in  the  medium  is 

%  =  Y  dynes, 
where  k  has  different  values  in  the  three  dielectrics. 


ELECTROSTATICS  37 

In  order  to  make  the  stresses  in  the  various  parts  of  the  insula- 
tion equal  it  would  be  necessary  to  place  next  to  the  conductor 
a  material  of  high  dielectric  constant  ki  and  to  decrease  this  con- 
stant gradually  in  succeeding  layers  in  inverse  proportion  to  the 
distance  from  the  center  of  the  conductor.  This  would  be  a 
very  expensive  process  and  it  is  not  necessary  since  good  results 
can  be  obtained  by  using  three  or  four  layers  of  dielectric. 

The  stress  in  the  outer  layer  is 

*-%>  (70) 

and  the  drop  of  potential  across  it  is 


f*R* 

=  JJ 


2g        #4 
the  stress  in  the  second  layer  is 


and  the  drop  of  potential  across  it  is 

2q  2q  ,      Rs  ,n  . 

r —  ctr  =  T~  log  -p— j  \i&) 

the  stress  in  the  inner  layer  is 


and  the  drop  of  potential  across  it  is 

Si  =  £B'gdr  =  |log|.  (75) 

The  difference  of  potential  between  the  conductor  and  the 
sheath  is 

^(7   i  xL2  ^w(7   i  -it^S  ^^/  -i  jTL4         /         \ 

The  capacity  of  the  cable  per  centimeter  length  is 


0.85  electro- 


2  /    log  —  +  -  log  —  +  -  loe  — ^         static  units. 
\6     g  0.4  T  4     g  0.6  T  2     S  1.27 


38  ELECTRICAL  ENGINEERING 

The  maximum  stress  or  gradient  in  (1)  is 

2q          2CE 

S^nax.  =  7  p    =   T   p    electrostatic  units, 


or  0max.  =  -j-g-  volts  per  centimeter;  (78) 

the  maximum  gradient  in  (2)  is 

O/^T? 
fiTmax    "-jjTfl-;  (79) 

and  the  maximum  gradient  in  (3)  is 

0max.   =  -j-g-.  (80) 

If  the  three  materials  have  the  same  dielectric  strength,  the 
thickness  should  be  so  chosen  that  the  three  maximum  stresses 
are  equal.  The  required  condition  is  that  JdRi  =  k2Rz  = 
ksRs.  In  the  present  case  the  three  materials  are  assumed  to 
have  the  same  maximum  allowable  stress  of  100,000  volts  per 
centimeter. 

If  the  allowable  maximum  stress  in  layer  (2)  had  been  only 
50,000  volts  per  centimeter  or  one-half  of  that  in  (1),  J?2  should 
have  been  chosen  so  that  k2R2  =  2kiRi  and  would  have  been 
1.2  instead  of  0.6. 

The  maximum  difference  of  potential  which  can  be  applied  to 
this  cable  can  be  found  by  substituting  numerical  values  in  any 
one  of  the  equations  for  maximum  gradient. 

From  equation  (78) 

=  2CE 

and  thus 

E  =  0max.  X  -^r1  =  100,000  X  s^r/rb  =  141,000  volts. 

^lU  £i    /\ 


The  gradient  from  conductor  to  sheath  is  shown  by  the  dis- 
continuous curve  in  Fig.  34(6),  and  the  area  under  the  curve 
represents  the  safe  operating  voltage  of  the  cable. 

In  this  case  by  grading  the  insulation  on  the  cable  without 
changing  the  outside  diameter  the  safe  operating  voltage  has 
been  increased  by  95  per  cent,  and  at  the  same  time  the  capacity 
has  been  decreased  about  45  per  cent. 


ELECTROSTATICS 


39 


42.  Air  Films  in  Generator  Slot  Insulation. — Fig.  35  shows  a 
section  of  one  slot  of  a  three-phase,  11, 000- volt  generator  with 
a  chain  winding.  The  insulation  on  the  coils  in  addition  to  the 
insulation  between  turns  consists  of  ti  =  0.2  cm.  of  molded  mica 
with  a  dielectric  constant  ki  =  4,  t%  =  0.4  cm.  of  varnished  cloth 
of  dielectric  constant  k2  =  5,  and  an  outer  wrapping  of  fiber  of 
thickness  £3  =  0.1  cm.  and  dielectric  constant  k3  =  2.  A  film 
of  air  of  thickness  £4  =  0.05  cm.  is  assumed  to  be  included  in  the 
slot.  Determine  the  maximum  stress  in  the  air. 


'Copper 

Double  Cotton  Cover 

(Neglected) 
Molded  Mica 
0.2  Cm.,  7ci  =  4 
Varnished  Cloth 
0.4  Cm.,/,;  2=  5 
Fibre  0.1  Cin.,&3>  2 


Core 


Film  0.05  Cm. ,fc 4-1 


FIG.  35. — Slot  insulation  for  an  11,000  volt  generator. 


The  maximum  voltage  to  neutral,  i.e.,  between  any  coil  and 
the  core,  is  found  at  the  terminal  coil  of  any  one  of  the  phases 

and  its  value  is  — ~-  X  \/2  =  8,960  volts  (see  Art.  123).     The 


side  of  the  coil  and  the  side  of  the  slot  may  be  considered  as  form- 
ing the  plates  of  a  condenser  with  four  layers  of  insulation  and 
the  stress  in  the  air  may  be  found  from  equation  (67), 


121 


8,960 


8,960 
0.23 


=  39,000  volts  per  centimeter. 


This  is  well  above  the  critical  stress  in  air  (31,000  volts  per 
centimeter)  and  the  air  will  therefore  break  down  and  ozone  and 
nitric  acid  will  form,  which  will  attack  the  insulation.  If  the  air 
is  continually  renewed  by  the  expanding  and  contracting  of  the 
coil  due  to  alternate  heating  and  cooling,  the  corrosive  action 
may  go  on  until  the  insulation  is  broken  down  and  the  coil  be- 
comes grounded.  The  corners  of  the  conductors  should  be 
rounded  off  to  prevent  local  concentration  of  flux  and  increased 
stress  as  shown  in  Fig.  35. 


40 


ELECTRICAL  ENGINEERING 


43.  Condenser  Bushing. — The  leads  from  transformers,  in 
some  cases  must  be  insulated  for  very  high  voltages  and  it  is  diffi- 
cult to  make  the  terminals  of  reasonable  size.  This  can  be  ac- 
complished only  by  making  each  part  of  the  insulating  material 
take  care  of  its  proper  proportion  of  the  total  stress.  One 
method  of  obtaining  this  result  is  illustrated  in  Fig.  36,  which  is  a 
cross-section  of  a  condenser  bushing. 


Insulating 
Compound 


Metal  Sheets 


IHHHHHh*- 

Equivalent  Circuit 


Transformer 
(Grounded) 


nductor 


FIG.  36. — Condenser  bushing. 


The  conductor  is  a  hollow  tube  of  sufficient  radius  to  reduce  the 
stress  at  its  surface  to  a  safe  value  and  it  is  placed  as  far  away 
from  the  other  terminal  as  possible.  The  tube  is  wrapped  with 
layers  of  insulating  material  separated  by  their  sheets  of  metal. 
Adjacent  metal  sheets  with  the  dielectric  between  them  form 
condensers  which  should  all  have  approximately  the  same  capac- 
ity, but  since,  due  to  the  increase  in  diameter,  the  outer  plates 


ELECTROSTATICS  41 

tend  to  have  greater  areas  than  the  inner  ones  they  must  be  made 
shorter.  The  lengths  of  succeeding  plates  are  reduced  by  equal 
amounts  in  order  to  get  as  long  a  surface  leakage  path  as  possible 
and  if  necessary  the  stresses  can  be  equalized  by  varying  the 
thickness  of  the  dielectric.  The  whole  bushing  is  covered  with  a 
cylinder  rilled  with  an  insulating  compound. 

44.  Dielectric  Strength. — The  dielectric  strength  of  an  insulat- 
ing material  is  measured  by  the  difference  of  potential  required  to 
puncture  it,  but  this  is  a  quantity  which  is  dependent  to  a  large 
extent  on  the  conditions  under  which  the  stress  is  applied.  The. 
test  is  usually  made  by  applying  an  alternating  voltage  to  the 
specimen,  which  is  placed  between  two  electrodes.  Frequencies 
of  from  40  to  60  cycles  per  second  are  generally  used  and  the 
shape  of  the  voltage  wave  is  approximately  sinusoidal.  In 
this  case  the  ratio  of  the  maximum  voltage,  which  is  responsible 
for  the  puncture,  to  the  effective  voltage,  which  is  indicated  by 
the  voltmeter,  is  \/2  to  1. 

The  dielectric  strength  is  variously  expressed  in  volts  per  mil, 
volts  per  inch,  volts  per  millimeter,  or  volts  per  centimeter. 

The  most  important  conditions  affecting  the  test  are:  (1) 
thickness  of  the  specimen;  (2)  shape  of  the  electrodes;  (3)  medium 
in  which  the  test  is  made;  (4)  temperature;  (5)  pressure;  (6)  time 
of  application  of  the  voltage;  (7)  frequency  of  the  impressed 
voltage;  (8)  wave  shape  of  the  impressed  voltage. 

1.  Fig.  37  shows  the  variation  of  apparent  dielectric  strength 
with  thickness. 

When  an  alternating  voltage  is  impressed,  the  changes  in  the 
direction  of  the  stress  cause  a  loss  in  the  dielectric  resulting  in 
a  rise  of  temperature,  which  decreases  the  dielectric  strength.  A 
thin  specimen  can  more  easily  radiate  this  heat  than  a  thicker  one 
and  so  its  temperature  does  not  rise  to  the  same  extent.  Further, 
it  is  more  difficult  to  make  thick  sheets  of  dielectric  as  homoge- 
neous and  free  from  flaws  as  thin  ones.  Both  of  these  con- 
ditions tend  to  make  the  thin  specimen  show  greater  strength 
under  test. 

When  a  considerable  thickness  of  dielectric  is  required  it  is 
advisable  to  build  it  up  of  a  number  of  thin  sheets  held  by  an 
insulating  varnish  as  this  makes  the  product  more  flexible 
and  it  is  not  likely  that  a  flaw  will  extend  through  more  than 
one  sheet. 

2.  If  the  electrodes  are  not  properly  shaped,  excessive  stresses 


42 


ELECTRICAL  ENGINEERING 


may  occur  at  certain  points  and  cause  a  breakdown  at  a  lower 
voltage  than  normal.  For  example,  transformer  oil  which  will 
stand  225  to  250  volts  per  mil  when  tested  between  half-inch 
discs  will  only  stand  about  200  volts  per  mil  when  tested  between 
half-inch  spheres. 

3.  If  a  sheet  of  material  is  tested  between  spherical  electrodes 
under  oil  it  will  show  a  lower  dielectric  strength  than  it  would  if 
tested  in  air,  since  the  air  will  break  down  first  and  so  give  the 
effect  of  flat  electrodes. 

4.  In  general  an  increase  of  temperature  results  in  a  decrease 
in  dielectric  strength  unless  the  increase  of  temperature  reduces 
the  amount    of   moisture  in  the  dielectric.     It  is  a  complex 
phenomenon. 


20       40       60       80      100    120 
Thickness  in  Mils. 


\ 

\ 

S^ 

20       40      60      80      100    120 
Time  in  Seconds 


FIG.  37. — Typical  curve  of  disruptive 
strength  vs.  thickness. 


FIG.  38. — Typical  curve  of  puncture 
voltage  vs.  time. 


5.  The  dielectric  strength  of  air  and  oil  increase  directly  with 
the  pressure. 

6.  Fig.  38   shows   the  variation  of  dielectric  strength  with 
time  of  application  of  the  voltage. 

A  given  material  will  withstand  a  much  higher  voltage  for  a 
short  time  than  it  will  for  an  extended  period.  For  this  reason,  in 
making  high-voltage  tests  on  machines  the  stress  must  be  applied 
for  a  specified  time. 

7.  Increase  of  frequency  results  in  an  increase  of  loss  and  a 
rise  in  temperature,  which  decreases  the  dielectric  strength. 

8.  The  breakdown  is  the  result  of  the  maximum  value  of  the 
impressed  voltage.     If  the  wave  is  peaked  the  ratio  of  maximum 
to  effective  value  will  be  greater  than  with  a  sine  wave  and  the 
material  will  break  down  with  an  effective  voltage  lower  than 
normal. 

In  the  following  table  are  given  approximate  values  of  the 
dielectric  constants  and  dielectric  strengths  for  materials  in  com- 
mon use  in  electrical  engineering. 


ELECTROSTATICS  43 

TABLE  OF  DIELECTRIC  CONSTANTS  AND  DIELECTRIC  STRENGTHS 


Material 

Dielectric 
constant 
ft 

Dielectric 
strength, 
volts  per  mil 

Remarks 

Air 

1  0 

79 

31  000  volts  per  cm 

Asphalt  

2  7 

30 

60  mils  thick 

Cellulose  acetate  

300-1  800 

Cloth   oiled 

750-1  000 

10  mils  thick 

Cloth,  varnished  

3  5-5  5 

500-1  300 

5-16  mils  thick 

Cotton  impregnated 

86 

7  mils  thick 

Cotton  not  impregnated 

21 

7  mils  thick 

Ebonite      

1.9-3.5 

1,700-3,750 

Up  to  20  mils 

Fiber 

2.0 

150-250 

Fullerboard 

7  5  at  100°C. 

Boiled  in  transformer 

Fullerboard  varnished  .  .  . 
Glass  

2.9  at    25°C. 
5.5-10.0 

300 
150-300 

oil. 

Gutta  percha 

3.0-5.0 

Mica 

2  5-6.0 

1,000-2,500 

Molded  mica  or  micanite. 
Paraffin  wax  .    . 

2.5-6.0 
1  9-2.3 

900 
300 

Porcelain 

4  0-6.0 

J300 

100  mils  thick 

Rubber     . 

2  0-3  O) 

\  220-240 

500  mils  thick 

Rubber  compounds  
Shellac  

3.0-4.0J 
2.75 

300-500 

Slate 

6  0-7  0 

5-10 

Transformer  oil  
Vacuum 

2.5 
0  9994 

225-250 
200 

Between  0.5-in.  discs 
0.2  in.  apart. 
Between    0.5-in. 
spheres    0.15-in. 
apart. 

45.  Breakdown. — When  a  discharge  takes  place  between  two 
electrodes  in  a  gaseous  or  liquid  dielectric,  the  natural  circulation 
of  the  material  heals  the  break  and  the  insulating  qualities  are 
not  impaired.  If  the  discharge  in  oil  is  very  heavy  some  of  the 
oil  becomes  carbonized  and  the  carbon  particles  tend  to  line  up 
in  the  field  of  greatest  stress  and  reduce  the  dielectric  strength. 

Particles  of  moisture  in  oil  due  to  their  high  dielectric  constant 
seek  the  most  intense  parts  of  the  field  and  cause  an  increase  of 
stress  and  tend  to  cause  breakdown. 

One-tenth  of  1  per  cent,  of  water  in  transformer  oil  will  reduce 
its  dielectric  strength  to  20  per  cent,  of  its  original  value. 


44  ELECTRICAL  ENGINEERING 

With  porcelain  or  glass  when  breakdown  occurs  the  material 
is  ruptured  and  its  insulating  value  destroyed. 

In  the  case  of  cotton  or  silk  fabrics  the  overstress  causes  local 
breakdown  accompanied  by  carbonization  and  a  weak  spot  de- 
velops which  gradually  extends  until  complete  rupture  occurs. 

46.  Dielectric  Losses. — If  a  constant  voltage  is  impressed  on 
the  terminals  of  a  condenser  (Fig.  13)  a  dielectric  field  is  set  up 
and  the  dielectric  material  is  in  a  state  of  stress.  So  long  as  the 
voltage  is  maintained  constant  there  is  not  consumption  of  energy 
due  to  this  cause.  When,  however,  an  alternating  voltage  is 
impressed  the  stresses  in  the  dielectric  alternate  in  direction.  If 
the  dielectric  were  perfectly  homogeneous  all  the  energy  stored 
in  the  field  during  the  increase  of  voltage  would  be  returned  as  it 
decreased  again  and  no  energy  would  be  consumed.  This  occurs 
in  the  case  of  gaseous  and  of  some  liquid  dielectrics.  In  solid 
dielectrics  some  energy  is  consumed  in  reversing  the  stress  in  the 
material  and  appears  as  heat  causing  a  rise  in  the  temperature  of 
the  dielectric.  The  loss  of  energy  is  proportional  to  the  square 
of  the  impressed  voltage,  that  is,  to  the  square  of  the  dielectric 
stress  and  is  called  the  dielectric  hysteresis  loss;  it,  however, 
differs  from  magnetic  hysteresis  in  that  it  causes  a  slight  lag  of 
flux  in  time  behind  the  voltage.  This  is  noticeable  only  at  high 
frequencies. 

A  further  energy  loss  occurs  in  dielectrics  due  to  the  leakage 
of  current  between  the  electrodes.  This  current  is  directly  pro- 
portional to  the  impressed  voltage  and  is  inversely  proportional 
to  the  resistance  of  the  dielectric.  The  resulting  loss  is  propor- 
tional to  the  square  of  the  current  and  therefore  to  the  square  of 
the  voltage.  It  occurs  with  both  direct  and  alternating  voltages. 

The  insulation  resistance  of  dielectrics  or  insulating  materials 
is  a  very  complex  quantity  and  varies  through  a  wide  range  with 
temperature  and  other  conditions.  Fig.  68  shows  the  variation 
of  the  resistance  of  slot  insulation  with  temperature.  The  pres- 
ence of  minute  quantities  of  moisture  in  the  dielectric  reduces 
the  insulation  resistance  to  very  low  values. 

There  is  no  direct  relation  between  the  insulation  resistance  of  a 
material  and  its  dielectric  strength.  For  instance,  dry  air  has  a 
very  high  insulation  resistance  but  very  low  dielectric  strength. 

Ordinarily  the  dielectric  losses  are  small,  but  in  the  case  of 
high-voltage  cables  where  the  stresses  are  very  high  they  may  be 
large  enough  to  cause  a  dangerous  rise  in  temperature  which  in 


ELECTROSTATICS  45 

turn  will  decrease  the  insulation  resistance  and  further  increase 
the  loss. 

47.  Surface  Leakage. — When  dielectric  surfaces  are  clean  and 
dry,  very  little  surface  leakage  takes  place  but  if  dust  or  mois- 
ture is  present  the  leakage  of  current  may  become  serious. 

The  flashing-over  of  clean  dielectric  surfaces  is  due  to  the 
breakdown  of  one  dielectric,  usually  air,  at  the  surface  of  a 
stronger  dielectric.  In  Fig.  28  leakage  occurs  over  the  long 
surface  acb  due  to  the  breaking  down  of  the  air  in  pockets  at  a 
and  6;  this  air  becomes  conducting.  The  shorter  surface  df  does 
not  exhibit  the  phenomenon  to  nearly  so  marked  an  extent. 

In  the  ordinary  pin-type  insulator,  discharges  over  the  surface 
take  place  due  to  the  breakdown  of  the  air  in  the  pockets  between 
the  petticoats  combined  with  the  presence  of  dust  and  moisture. 

48.  Corona. — When  the  electrostatic  stress  or  potential  gradi- 
ent at  any  point  in  air  exceeds  about  30,000  volts  per  centimeter 
a  brush  discharge  takes  place  and  the  air  becomes  conducting 
and  luminous.     The  discharge  does  not  necessarily  extend  from 
the  positive  to  the  negative  electrode  but  exists  only  in  the  region 
where  the  dielectric  strength  of  the  air  has  been  exceeded. 

In  the  case  of  two  parallel  wires  suspended  in  air  the  discharge 
or  corona  first  appears  at  any  rough  points  on  the  wires  and  fin- 
ally forms  a  luminous  envelope  about  them  which  increases  in 
diameter  as  the  voltage  is  raised. 

If  the  increase  in  the  effective  diameter  of  the  wire  due  to  the 
presence  of  the  corona  results  in  a  decrease  of  the  stress  a  flash- 
over  between  the  wires  will  not  result;  but  if  the  increase  in  di- 
ameter so  decreases  the  distance  between  the  conductors  that  the 
stress  at  the  surface  of  the  conducting  envelope  is  not  reduced 
below  the  breakdown  stress  of  the  air,  the  diameter  of  the  envel- 
ope increases  until  a  flash  takes  place  from  one  wire  to  the  other. 

Corona  is  accompanied  by  a  loss  of  energy  proportional  to  the 
square  of  the  voltage  rise  above  the  critical  voltage  at  which  the 
discharge  begins. 


CHAPTER  II 


MAGNETISM  AND  ELECTROMAGNETICS 

49.  Magnetization. — When  bodies  are  magnetized  magnetic 
forces  act  at  every  point  throughout  their  volume  and  lines  of 
magnetic  induction  pass  through  them.  There  are  two  kinds  of 
magnetic  poles  just  as  there  are  two  kinds  of  electric  charges; 
as  a  positive  electric  charge  appears  where  a  dielectric  flux  leaves 
a  surface,  so  a  positive  magnetic  pole  appears  where  a  magnetic 
flux  leaves  a  surface.  The  positive  magnetic  pole  is  called  a 
north  pole.  Similarly  a  negative  magnetic  pole  or  south  pole 
appears  where  a  magnetic  flux  enters  a  surface.  - 


FIG.  39. — Magnet. 


FIG.  40. — Magnet  with  armature  on. 


Thus  a  body  which  is  magnetized  has  a  north  pole  at  one  part 
of  its  surface  and  an  equal  south  pole  at  another  part  unless  the 
magnetic  path  forms  a  closed  circuit  as  in  Fig.  40. 

Fig.  39  represents  a  horseshoe  magnet.  The  lines  of  magnetic 
induction  pass  through  it  in  the  direction  shown,  leaving  the  sur- 
face at  N  and  entering  it  again  at  S.  Thus  TV  is  a  positive  mag- 
netic pole  or  a  north  pole  and  S  is  a  negative  magnetic  pole  or  a 
south  pole. 

Fig.  40  represents  the  same  magnet  with  its  armature  on.  The 
lines  of  magnetic  induction  pass  in  the  same  direction  as  before, 
but  the  circuit  is  closed  and  the  poles  do  nor  appear  until  a  gap 
is  made  by  removing  the  armature, 

46 


MAGNETISM  AND  ELECTROMAGNETICS         47 

60.  Laws  of  Magnetism.  —  First  Law.  —  Like  magnetic  poles 
repel  one  another;  unlike  magnetic  poles  attract  one  another. 

Second  Law.  —  The  force  exerted  between  two  magnetic  poles 
is  proportional  to  the  product  of  their  strengths  and  is  inversely 
proportional  to  the  square  of  the  distance  between  them.  This 
law  can  be  expressed  by  the  formula 


where  m  and  mi  are  the  pole  strengths,  r  is  the  distance  between 
them  in  centimeters,  and  /  is  the  force  exerted  between  them  in 
dynes.  If  m  and  mi  are  like  poles  the  force  is  a  repulsion  and 
/  is  positive. 

The  unit  of  pole  strength  is  denned  as  follows:  A  magnetic 
pole  has  unit  strength  if,  when  placed  at  a  distance  of  1  cm. 
from  a  similar  pole,  it  repels  it  with  a  force  of  one  dyne. 

The  force  exerted  on  a  unit  pole  at  a  distance  of  r  cm.  from  a 
pole  of  strength  m  is 

/  =  ™2  dynes.  (82) 

51.  Magnetic  Field.  —  The  space  surrounding  a  magnetic  pole 
or  a  current  of  electricity  in  which  magnetic  forces  act  is  called 
a  magnetic  field.     The  direction  of  the  force  at  any  point  in  the 
field  is  the  direction  in  which  a  unit  north  pole  placed  at  the  point 
would  tend  to  move  and  its  intensity  is  the  force  in  dynes  exerted 
on  the  unit  pole. 

The  magnetic  field  is  represented  by  lines  of  magnetic  induc- 
tion or  magnetic  flux  drawn  in  the  direction  of  the  force. 

Unit  magnetic  force  or  unit  magnetizing  force  produces  one 
line  of  magnetic  flux  per  square  centimeter  in  air  and  p  lines  per 
square  centimeter  in  a  magnetic  material  of  permeability  /z. 

The  magnetic  force  at  a  point  is  expressed  in  dynes  and  is 
represented  by  3C;  the  magnetic  flux  density  at  a  point  is  ex- 
pressed in  lines  per  square  centimeter  and  is  represented  by  (B. 

Fig.  41  shows  the  magnetic  fields  produced  in  certain  cases. 
The  lines  of  induction  are  all  closed  lines  and  extend  from  a 
north  to  a  south  pole  in  air  and  from  a  south  to  a  north  pole  in- 
side the  magnetic  material  or  the  generator  of  m.m.f. 

52.  Magnetic  Flux.—  The  total  number  of  lines  of  magnetic 
induction  passing  through  a  given  section  is  called  the  magnetic 
flux  through  the  section  and  is  represented  by  <£. 


48 


ELECTRICAL  ENGINEERING 


The  unit  of  magnetic  flux,  which  is  one  line,  is  called  the  max- 
well. 

53.  Flux  from  Unit  Pole. — At  every  point  on  a  sphere  of  1  cm. 
radius,  surrounding  a  unit  pole  as  center,  a  similar,  unit  pole  is 
repelled  with  a  force  of  one  dyne.  There  must  therefore  be  one 
line  of  induction  per  square  centimeter  passing  through  the 
surface,  and  since  the  surface  is  4?r  sq.  cm.  the  total  flux  from  the 
unit  pole  is 

<I>  =  4rr  lines. 

The  flux  from  a  pole  of  strength  m  is 
<1>  =  4rrm  lines. 


(a)  Field  of  a  Bar  Magnet 


(  b  )   Field  of  a  Solenoid 


(c)  Field  near  arilsolated  North  Pole 


(  d)  Field  of  a  Dynamo 


FIG.  41. — Magnetic  fields. 

Thus  a  unit  north  pole  is  associated  with  each  4?r  lines  leaving 
a  surface  and  a  unit  south  pole  with  each  4?r  lines  entering  a 
surface. 

64.  Magnetic  Potential. — The  magnetic  potential  of  an  iso- 
lated magnetic  pole  is  the  work  done  in  carrying  a  unit  north 
pole  from  an  infinite  distance  to  the  point  against  the  forces  in 
the  magnetic  field. 

Fig.  42  shows  a  north  magnetic  pole  of,  strength  m.     Its  field 


MAGNETISM  AND  ELECTROMAGNETICS         49 

extends  out  radially  in  all  directions  and  the  magnetic  force  at  a 
distance  of  r  cm.  from  m  is 

m  , 
5C  =  -2  dynes. 

The  magnetic  potential  of  the  point  P  at  a  distance  of  ri  cm. 
from  m  is  the  work  done  in  carrying  a  unit  north  pole  from  an 
infinite  distance  to  the  point  against  the  force  of  repulsion  of  m. 
The  work  done  is 


Jjr       f«       ,        f"    dr       m 

W  =        OC  dr  =        m-2  =  - 

Jri  Jri  i 


Therefore  the  magnetic  potential  of  a  point  at  a  distance  of  ri  cm. 
from  an  isolated  magnetic  pole  of  strength  m  is 


(83) 


FIG.  42. — Magnetic  potential. 

The  difference  of  magnetic  potential  between  the  points  P  and 
Q  in  Fig.  42  is 

•MT       Cr\n  j         fr'     dr       m       m 

M  =  I    5C  dr  =  I    m  -T  = 

Jr,  Jr,        ^         ri         r2 

Thus  the  difference  of  magnetic  potential  between  two  points  is 
the  line  integral  of  the  magnetic  force  between  them. 

55.  Magnetomotive -Force. — The  difference  of  magnetic  poten- 
tial or  the  line  integral  of  the  magnetic  force  between  two  points 
is  called  the  magnetomotive  force  (m.m.f.)  between  the  points. 
It  causes  magnetic  flux  to  pass  from  one  to  the  other. 

Unit  m.m.f.  will  produce  one  line  of  magnetic  flux  through  a 
(cm.)3  of  air  and  /*  lines  through  a  (cm.)3  of  a  magnetic  material 
of  permeability  /*.  It  is  called  the  gilbert. 

Since  m.m.f.  is  the  line  integral  of  the  magnetic  force  or 


M  =    I  5C  dr, 


50        ELECTRICAL  ENGINEERING 

therefore  the  magnetic  force  is  the  space  rate  of  change  of  m.m.f. 
or 

30  =  w'  (84) 

and  thus  magnetic  force  is  the  m.m.f.  per  centimeter. 

56.  Permeability.  —  Permeability  is  the  ratio  of  the  magnetic 
conductivity  of  a  substance  to  the  magnetic  conductivity  of  air 
and  is  represented  by  /*. 

Lines  of  magnetic  flux  pass  through  air  or  any  other  substance 
except  iron,  nickel  or  cobalt,  as  easily  as  they  do  through  a 
vacuum.  The  permeability  of  such  substances  is  for  all  practical 
purposes  the  same  and  is  taken  as  unity.  Iron  and  its  compounds 
and  to  a  lesser  degree  nickel  and  cobalt  are  found  to  allow  mag- 
netic lines  to  pass  through  them  much  more  easily  than  empty 
space;  that  is,  a  giveHm.m.f.  will  produce  a  much  larger  flux 
through  a  volume  of  iron  than  it  will  through  an  equal  volume  of 
air.  The  permeability  of  the  iron  is  therefore  greater  than  that 
of  the  air  and  is  expressed  by  some  number  greater  than  unity. 

The  permeability  of  magnetic  materials  is  not  constant  but 
varies  with  the  induction  density  as  shown  in  Art.  72. 

67.  Magnetic  Reluctance.  —  The  reluctance  of  a  magnetic 
circuit  may  be  defined  as  the  resistance  offered  by  the  circuit  to 
the  passage  of  magnetic  flux  through  it  and  is  represented  by  (R. 

If  a  m.m.f.  M  is  applied  to  a  path  of  length  I  cm.,  of  uniform 
section  A  sq.  cm.  and  of  permeability  /*,  the  m.m.f.  per  centimeter 

will  be  -y  and  this  will  produce  through  each  square  centimeter  a 

M 
flux  or  flux  density  <B  =  -y-  n  lines  and  the  total  flux  through  the 

path  will  be 

M  MM 


Thus  the  reluctance  of  a  path  of  uniform  section  is 

(R  =  4~' 
Ap 

and  is  directly  proportional  to  its  length  and  inversely  propor- 
tional to  its  sectional  area  and  to  the  permeability  of  the  material 
forming  it. 

The  equation  connecting  the  m.m.f.  acting  on  a  path,  the  re- 


MAGNETISM  AND  ELECTROMAGNETICS         51 

luctance  of  the  path  and  the  flux  through  the  path  can  be  written 
in  three  ways: 

1.  *  =  £  (86) 

01 

the  flux  is  equal  to  the  m.m.f.  divided  by  the  reluctance; 

2.  M  =  $(R  (87) 
the  m.m.f.  is  equal  to  the  flux  jnultiplied  by  the  reluctance; 

3.  (R  =  f-v  (88> 

the  reluctance  is  equal  to  the  m.m.f.  divided  by  the  flux. 

Assuming  the  flux  to  be  unity  in  the  last  equation  the  reluc- 
tance of  the  path  may  be  defined  as  the  m.m.f.  required  to  pro- 
duce unit  flux  through  it. 

58.  Permeance.  —  The  permeance  of  a  magnetic  path  is  the 
reciprocal  of  its  reluctance  and  is  represented  by  (P;  thus 


and  assuming  that  the  m.m.f.  acting  is  unity,  the  permeance 
may  be  defined  as  the  flux  produced  through  the  path  by  unit 
m.m.f. 

The  permeance  of  a  path  of  uniform  section  is 


and  is  directly  proportional  to  the  sectional  area  and  to  the  per- 
meability, and  is  inversely  proportional  to  the  length  of  the  path. 

59.  Electromagnetics.  —  The  region  surrounding  a  conductor 
carrying  a  current  of  electricity  is  a  magnetic  field.  A  current 
of  electricity  therefore  represents  a  m.m.f.  If  the  conductor 
is  isolated  from  other  magnetic  forces  the  lines  of  force  will  form 
circles  around  it. 

Maxwell's  Corkscrew  Rule.  —  The  direction  of  the  current  and 
that  of  the  resulting  magnetic  force  are  related  to  one  another 
as  the  forward  travel  and  the  twist  of  an  ordinary  corkscrew. 
This  rule  is  illustrated  in  Fig.  43. 

The  symbol  ©  represents  a  current  flowing  down  and  O  a 
current  flowing  up. 


52  ELECTRICAL  ENGINEERING 

Faraday  discovered  that  a  current  is  induced  in  a  closed  coil 
of  wire  when  a  magnet  is  brought  near  it.  The  same  effect  is 
noticeable  if  a  coil  of  wire  carrying  current  is  moved  to  or  from 
the  closed  coil,  or  if  the  second  coil  is  fixed  in  position  and  the 
current  in  it  is  varied.  The  induced  current  only  exists  while 
the  magnet  or  inducing  coil  is  moving  with  respect  to  the  fixed 
coil  or  while  the  current  in  the  inducing  coil  is  varying. 


FIG.  43. — Magnetic  flux  produced  by  an  electric  current. 


The  induced  current  is  due  to  the  fact  that  a  difference  of 
potential  or  electromotive  force  is  produced  in  the  circuit  by 
changing  the  number  of  lines  of  magnetic  flux  threading  through 
it  or  by  causing  lines  of  magnetic  flux  to  cut  across  it. 

60.  Laws  of  Induction. — First  Law. — A  change  in  the  number 
of  lines  which  pass  through  a  closed  circuit  induces  a  current 
around  the  circuit  in  such  a  direction  as  to  oppose  the  change  in 
the  flux  threading  the  circuit. 

Second  Law. — The  electromotive  force  induced  around  a  closed 
circuit  is  equal  to  the  rate  of  change  of  the  flux  which  passes 
through  the  circuit;  or  the  electromotive  force  induced  in  a  con- 
ductor is  equal  to  the  rate  at  which  it  cuts  across  lines  of  magnetic 
flux. 

61.  Unit  of  Electromotive  Force. — The  absolute  unit  of  elec- 
tromotive force  (e.m.f.)  is  the  e.m.f.  induced  in  a  coil  of  one  turn 
when  the  flux  threading  the  coil  is  changing  at  the  rate  of  one 
line  per  second;  or  it  is  the  e.m.f.  induced  in  a  conductor  when 
it  is  cutting  one  line  per  second.     It  is  called  the  abvolt. 

The  practical  unit  is  the  e.m.f.  produced  by  cutting  108  lines 
per  second  and  is  called  the  volt.  E.m.f.  is  commonly  called 
voltage. 

To  change  from  absolute  units  of  e.m.f.  to  volts  divide  by  108. 


MAGNETISM  AND  ELECTROMAGNETICS         53 

If  a  coil  of  wire  has  n  turns  and  the  flux  through  it  is  changing 
at  the  rate  "-?  lines  per  second,  the  e.m.f.  induced  in  the  coil  is 

e  =  —  n  -£  absolute  units.  (SI) 

at 

The  negative  sign  is  used  because  when  the  flux  is  decreasing 
the  induced  e.m.f.  is  in  the  positive  direction,  that  is,  it  tends  to 
prevent  the  decrease  of  the  flux. 

62.  Force  Exerted  by  a  Magnetic  Field  on  an  Electric  Cir- 
cuit.— Every  part  of  an  electric  circuit  situated  in  a  magnetic 
field  is  acted  upon  by  a  force  at  right  angles  both  to  the  direction 
of  the  current  and  to  the  lines  of  force,  and  the  circuit  as  a 
whole  is  acted  upon  by  forces  tending  to  move  it  into  the  position 
where  it  will  include  the  greatest  possible  flux. 


Force  Force=fflil    Force- ©JI  Cos  0 

(a)  (b)  (c)  (d)  (e) 

FIG.  44. — Force  on  an  electric  conductor  in  a  magnetic  field. 

In  Fig.  44,  (a)  represents  a  uniform  magnetic  field  between  two 
unlike  poles  and  (6)  represents  the  field  surrounding  a  conductor 
carrying  current.  If  the  conductor  is  placed  in  the  uniform  field 
the  resultant  distribution  will  be  as  shown  in  (c).  The  intensity 
of  the  field  above  the  conductor  will  be  greater  than  that  below 
and  a  force  /  will  act  on  the  conductor  at  right  angles  to  it  and  to 
the  lines  of  flux,  tending  to  push  it  out  of  the  field.  This  force  is 
directly  prop6rtional  to  the  intensity  of  the  field  or  the  flux  den- 
sity, to  the  length  of  the  conductor  in  the  field  and  to  the  strength 
of  the  current,  or 

/  =  (&II  dynes,  (92) 

where  <B  is  the  flux  density  in  lines  per  square  centimeter, 
I  is  the  length  of  the  conductor  in  centimeters, 
/  is  the  strength  of  the  current  in  absolute  units. 
It  has  been  assumed  that  the  conductor  lies  at  right  angles  to 
the  lines  of  force  as  shown  at  (d) ;  if  it  is  inclined  at  an  angle  6  to 
this  direction  as  at  (e)  then  the  force  is  reduced  to 

/  =  (Btf  cos  6.  .(93) 


54  ELECTRICAL  ENGINEERING 

In  Fig.  45 (a),  AB  is  a  section  through  a  coil  of  wire  carrying 
current.  A  is  acted  upon  by  a  force  fA  tending  to  move  it  down 
and  B  is  acted  upon  by  a  smaller  force  fB)  tending  to  move  it  up. 
The  coil  will  move  down  until  A  and  B  are  in  fields  of  equal 
strength  and  the  flux  threading  the  coil  is  maximum.  The  flux 
produced  by  the  current  in  the  coil  is  in  the  same  direction  as 
the  main  field.  (6)  shows  the  same  coil  reversed  and  in  this  case 
it  will  be  forced  out  of  the  field,  (c)  shows  the  coil  near  the  posi- 
tion of  maximum  inclosure  of  flux  but  with  the  flux  produced  by 
the  current  opposing  the  main  flux.  The  coil  is  forced  around 
to  the  position  A'B1 '. 


00  (O  («) 

FIG.  45. — Force  exerted  by  a  magnetic  field  on  a  coil  of  wire. 

63.  Unit  Current. — If  a  conductor  carrying  one  absolute  unit 
of  current  is  placed  in  a  magnetic  field  of  unit  strength  at  right 
angles  to  the  lines  of  force,  each  centimeter  of  its  length  will  be 
acted  upon  by  a  force  of  one  dyne.     This  absolute  unit  of  current 
is  called  the  absampere. 

The  practical  unit  is  one-tenth  of  the  absolute  unit  and  is 
called  the  ampere. 

64.  Transformation  of  Mechanical  Energy  to  Electrical  Energy. 
— In  Fig.  46,  if  the  conductor  is  moved  through  a  distance 
dx  at  right  angles  to  the  flux  against  the  force  /,  the  work  done  is 

dw  =  f  dx  =  (BZJ  dx  =  (&l  dx  7, 

but  (&l  dx  is  the  flux  cut  in  moving  through  the  distance  dx  and  is 
=  d<j>,  therefore 

*  dw  =  /  d<j>,  (94) 

and  the  work  done  in  moving  a  current  across  a  magnetic  field  is 
equal  to  the  product  of  the  current  and  the  flux  cut. 

In  moving  completely  across  the  pole  face  the  work  done  is 

W  =    f /  d<f>  =  /*, 
where  <i>  is  the  flux  from  the  pole. 


MAGNETISM  AND  ELECTROMAGNETICS         55 

If  I  is  expressed  in  absolute  units  and  $  in  maxwells,  W  is  in 
ergs. 

If  the  motion  through  the  distance  dx  takes  place  in  time  dt  sec., 
the  work  done  is 

dw  =  I  d<j>  =  I  dfifc  =  el  dt,  (95) 

where  e  =  -jr  is  the  e.m.f.  generated  in  the  conductor  and  I  dt 

=  dq  is  the  quantity  of  electricity  raised  through  the  difference 
of  potential  e.  Therefore,  the  mechanical  work  supplied  to 
move  the  conductor  through  the  distance  dx  against  the  force  / 
is  used  up  in  doing  the  electrical  work  of  raising  a  quantity  of 


FIQ.  46. — Transformation  of  mechanical  energy  to  electrical  energy. 

electricity  dq  through  a  difference  of  potential  e,  or  in  driving 
a  current  /  against  an  e.m.f.  e  for  a  time  dt.  Thus  mechanical 
energy  is  transformed  into  electrical  energy.  This  is  what  takes 
place  in  an  electric  generator. 

If  electric  power  is  supplied  to  drive  the  current  /  against  the 

e.m.f.  e  —  -j-,  for  a  time  dt  energy  is  supplied 
dw  =  eldt  =  ftldt  =  Id<t>- 

the  conductor  exerts  a  force  /  =  (BZ I  dynes  through  a  distance  dx 
and  does  mechanical  work, 

/  dx  =  &II  dx  =  /  d<t>. 

This  is  the  action  of  an  electric  motor. 

65.  Electric  Power  and  Energy. — When  an  electric  circuit 
carrying  a  current  /  absamp.  incloses  a  flux  <f>,  which  is  not  pro- 


56  ELECTRICAL  ENGINEERING 

duced  by  the  current,  the  potential  electrical  energy  of  the  system 
is 

w  =  <t>I  ergs.  (96) 

This  amount  of  energy  must  have  been  expended  in  bringing  the 
electric  circuit  into  its  position. 

Power  is  the  rate  of  flow  or  the  rate  of  transformation  of 
energy.  The  mechanical  power  required  to  move  the  con- 
ductor AB  across  the  field  and  thus  to  vary  the  flux  <£  is 

n       dw       ,  d<l>         .  r  dx       ,    , 

P  =  ~Ji  =  I  -T7  =  (&II  -r  =  fv  dyne-cm,  per  second, 

where  v  =  ~r  is  the  velocity  of  the  conductor  normal  to  the  field. 
The  electric  power  generated  during  the  motion  is 

D       dw       rd<l>         j 

P  =  ~j~  =  I  -=-  —el  ergs  per  second, 

where  e  =  -r  is  the  e.m.f .  generated  in  the  conductor  by  cutting 

the  flux. 

The  electric  power  in  a  circuit  is  the  product  of  the  current  and 
the  e.m.f.  in  the  circuit. 

The  practical  unit  of  power  is  the  watt.  It  is  the  power  in  a 
circuit  carrying  one  ampere  when  the  e.m.f.  across  it  is  one  volt. 

•P Watts   =   ^Volts   X  J  Amperes.  (97) 

The  kilowatt,  which  is  1,000  watts,  is  more  commonly  used 
where  the  amounts  of  power  are  large.  One  horsepower  is 
equivalent  to  746  watts. 

The  electric  energy  transformed  in  a  circuit  is  the  product  of 
the  power  and  the  time.  The  practical  units  of  electric  energy 
are  the  watt-second  or  joule,  the  watt-hour  and  the  kilowatt-hour. 

The  absolute  unit  of  electric  energy  is  the  erg. 

1  watt-sec.  =  1  volt  X  1  amp.  X  1  sec. 

=  1  abvolt  X  10  8X  1  absamp.  X  10"1  X  1  sec. 
-«  107  ergs. 

66.  Intensity  of  Magnetic  Fields  Produced  by  Electric  Currents. — 
The  following  cases  are  of  special  importance:  (A)  At  the  center  of  a 
circular  loop  of  wire  carrying  a  current  I  absolute  units  (Fig.  47). 


MAGNETISM  AND  ELECTROMAGNETICS          57 

If  3C  is  the  field  intensity  at  0,  the  center  of  the  loop,  a  magnetic  pole 
of  strength  m  placed  at  this  point  will  be  acted  upon  by  a  force 

/  =  w3C  dynes 

in  a  direction  perpendicular  to  the  plane  of  the  coil. 
The  pole  will  produce  at  the  wire  a  field  of  intensity 

^i  =        dynes, 


and  a  flux  density 

where  r  cm.  is  the  radius  of  the  loop. 


m  ,. 
=  -^  lines  per  square  centimeter, 


Flux 


FIG.  47.  —  Intensity  of  the  magnetic  field  at  the  center  of  a  circular  coil  of  wire. 

This  field  will  act  on  the  wire  with  a  force 
/i  =  (BZ7  dynes 

in  a  direction  perpendicular  to  the  plane  of  the  coil,  where  I  =  2-nr  is 
the  length  of  the  wire  in  centimeters. 
Substituting  the  values  of  (B  and  I  gives 

/i  =       X  2irrl  =  m  --  dynes, 


but  the  forces  /  and  f\  are  equal  and  therefore 
and 


27T/ 

m3C  =  m. > 


2irl 
3C  =  —  dynes.  (98) 

The  flux  density  at  the  center  of  the  loop  is 

27T/ 

(B  =  3C  =  —  lines  per  square  centimeter. 

If  /  is  expressed  in  amperes  the  field  intensity  at  the  center  of  the 
loop  is 

dynes.  (99) 


58 


ELECTRICAL  ENGINEERING 


(B)  At  a  distance  of  r  cm.  from  a  long  straight  wire  carrying  a  current 
of  7  absolute  units  (Fig.  48). 

If  3C  is  the  intensity  of  the  field,  the  work  done  in  moving  a  unit  mag- 
netic pole  around  the  wire  at  a  distance  r  cm.  from  it  against  the  force 
JC  is 

w  =  2irr3£  ergs. 

The  work  done  is  equal  to  the  product  of  the  current  and  the  flux  cut 
and,  therefore, 

2xr3C  =  4x7, 
and 


21 

3C  =  —  dynes; 


(100) 


the  intensity  of  the  field  varies  directly  as  the  strength  of  the  current 
and  inversely  as  the  distance  from  the  wire. 
The  flux  density  at  distance  r  is 


27 

(B  =  3C  =  —  lines  per  square  centimeter. 


If  7  is  in  amperes  the  field  intensity  is 

0  27 

-y-  dynes, 


3C 


and  the  flux  density  is 


(B  = 


0.27 


lines  per  square  centimeter. 


(101) 


(102) 


(103) 


Unit 
Pole 


Motion 


Unit  current  would  produce  a  flux  density 
of  2  lines  per  square  centimeter  at  a  distance 
of  1  cm.  from  a  straight  wire  and  it  would 
produce  this  flux  through  a  distance  of  2ir 
cm.,  or  it  would  produce  a  flux  density  of  4x 
lines  per  square  centimeter  through  a  dis- 
tance of  1  cm.  in  air.  Thus  one  absolute 
unit  of  current  represents  a  m.m.f.  of  4*- 
gilberts  and  one  ampere  represents  a  m.m.f. 
of  0.47T  gilberts. 

(C)  Between  two  parallel  wires  A  and  B, 
Fig.  49,  at  a  distance  of  D  cm.  apart  and 
carrying  equal  currents  7  but  in  opposite 
directions. 

The  field  intensity  or  magnetic  force  at 

P°int  P  .distot  *  Cm"  frOm  A  and  D  .~  '  Cm' 
from  B  is  the  resultant  of  the  magnetic  forces 

due  to  the  currents  in  A  and  B.     Since  these 

forces  act  in  the  same  direction  at  all  points  between  the  wires  they 
can  be  added  directly;  the  field  intensity  is  by  equation  100 


straight  wire. 


3C  = 


27 


-f 


27 

D  -r 


dynes, 


(104) 


MAGNETISM  AND  ELECTROMAGNETICS         59 

and  the  flux  density  at  P  is 

27         27 

(B  =  3C  = h  £>  _    lines  per  square  centimeter.  (105) 

\  (D)  At  any  point  on  the  axis  of  a  short  coil  of  radius  r  cm.  (Fig.  50). 


n  =  number  of  turns  in  the  coil, 

7  =  current  in  the  coil  in  absolute  units. 


FIG.  49. — Magnetic  field  between  two  parallel  wires. 

Take  any  point  P  on  the  axis  at  a  distance  x  cm.  from  the  plane  of  the 
coil  and  let  3C  be  the  field  intensity  there.  If  a  pole  of  strength  m  is 
placed  at  P  it  will  be  acted  on  by  a  force  of  w3C  dynes  perpendicular  to 
the  plane  of  the  coil.  The  force  exerted  on  the  coil  by  the  pole  is  equal 
and  opposite  to  the  force  exerted  on  the  pole  by  the  coil. 


//cos  e 


i®00©0©©0©©0©©to©©0©000©0©©0B 

D 

FIG.  50. — Intensity  of  the  magnetic       FIG.  51. — Magnetic  field  in  a  solenoid, 
field  on  the  axis  of  a  short  coil. 

The  field  intensity  at  the  wire  due  to  the  pole  m  at  P  is  -^  dynes  and 

the  flux  density  is  -™  lines  per  square  centimeter,  where  d  cm.  is  the  dis- 
tance from  the  point  to  the  wire;  the  length  of  wire  is  2irrn  cm.  and  there- 
fore the  force  exerted  on  it  is 

/  =  ^  X  2  irrnl  dynes. 

This  force  acts  at  right  angles  to  the  lines  of  flux  and  may  be  resolved 
into  two  components,  /  cos  e  in  the  plane  of  the  coil  and  /  sin  6  perpen- 


60  ELECTRICAL  ENGINEERING 

dicular  to  the  plane  of  the  coil.  The  component  /  cos  6  taken  around 
the  loop  is  zero  and  therefore  the  component  /  sin  6  is  equal  in  magnitude 
to  the  force  w3C. 

The  field  intensity  or  magnetic  force  at  P  is,  therefore, 

/sinfl      2irrnl    . 
3C  =  -  =  —  ^  —  sin  e 
m  dz 

r  2irr2nl 


snce 


V2  +  x2 

(E)  On  the  axis  of  a  long  solenoid.    In  Fig.  51  AB  is  a  solenoid  of 
length  I  cm.  and  radius  r  cm.     If  n  is  the  number  of  turns  in  the  sole- 

noid, the  number  of  turns  in  the  section  CD  of  width  dx  is  j  dx. 
The  field  intensity  at  P  due  to  the  section  CD  is,  by  equation  (106), 


where  /  is  the  current  in  the  solenoid. 
The  field  intensity  due  to  the  complete  solenoid  is 


J.1 
2     27iT2/          n 
l(r.  +  irl)KX    j    d*. 
•o 


'2 

To  integrate  this  let  angle  CPT  =  6,  then 

x  =  r  cot  0, 
(r2  +  z2)%  =  r3(l  +  cot2  o)*  =  r3  cosec3  6  =     ~ 

rfx  =  -  r  cosec2  d  de  =  - 


if  the  solenoid  is  assumed  to  be  very  long  the  limits  may  be  taken 
as  0  and  TT,  and,  therefore, 


27rn7  f°  . 
=  --  —        sm 

2irn/  .         10 

=  -      -    [COS  e\l 


dynes.  (107) 


If  the  current  is  expressed  in  amperes 


0.47Ttt7     ,  /ino\ 

OC  =  —  j  —  dynes,  (108) 


MAGNETISM  AND  ELECTROMAGNETICS          61 

and  the  field  intensity  on  the  axip  of  a  long  solenoid  is  proportional  to 
the  product  of  amperes  and  turns  or  ampere-turns  and  is  inversely  pro- 
portional to  the  length  of  the  solenoid. 

The  field  intensity  throughout  the  volume  enclosed  by  the  solenoid 
is  practically  uniform  except  near  the  ends  and  can  be  expressed  by 
equation  108. 

67.    Magnetomotive    Force    of    a 
Solenoid. — The  m.m.f.  of  a  solenoid 
is  the  line  integral  of  the  magnetic  / 
forces  along  any  closed  path  through  \ 
it  and  is  measured  by  the  work  done 
in     carrying    a    unit    magnetic    pole 
around  the  closed  path  (Fig.  52).  FlG -  52.— Magnetomotive 

.  ,  force  of  a  solenoid. 

The   work    done   is    equal  to  the 
product  of  the  current  and  the  flux  cut,  and  thus 

m.m.f.  =  4irnl,  where  /  is  in  absolute  units, 
or 

m.m.f.  =  OAirnl,  where  /  is  in  amperes. 

The  m.m.f.  is  proportional  to  the  ampere-turns  of  the  coil.  It 
does  not  make  any  difference  whether  it  is  a  small  current  in  a 
large  number  of  turns  or  a  large  current  in  a  small  number  of 
turns. 

The  m.m.f.  of  one  ampere  in  one  turn  or  one  ampere-turn  is 
0.4T  gilberts. 

In  studying  the  characteristics  of  electrical  machinery  it  is 
more  convenient  to  use  the  ampere-turn  as  the  unit  of  m.m.f. 
Thus  the  m.m.f.  of  a  field  coil  of  n  turns  carrying  a  current  / 
amp.  is  specified  as  nl  ampere-turns  instead  of  QAirnl  gilberts. 

68.  Examples. — 1.  The  solenoid  in  Fig.  53  has  nl  ampere-turns  and  is 
wound  on  a  ring  of  non-magnetic  material.  The  m.m.f.  of  the  solenoid  is 
M  =  QAirnl  gilberts. 

If  I  cm.  is  the  mean  length  of  the  path  and  A  sq.  cm.  is  the  sectional  area 
of  the  path,  the  reluctance  is 

~~  A' 
and  the  flux  inside  the  ring  is 

_  M  _  0.47m/,. 

The  flux  density  in  the  ring  is 

(B  —  -j  lines  per  square  centimeter, 


62 


and  the  magnetizing  force  is 


ELECTRICAL  ENGINEERING 

4>       0.4rm/ 


and  is  the  m.m.f.  per  centimeter. 

If  the  solenoid  is  wound  on  an  iron  ring  of  permeability  M,  the  reluctance 
is  reduced  and  becomes 

(Ri  =  --, 
the  flux  is  increased  to 


and  the  flux  denaty  or  induction  density  is  increased  to 

CBI=-; 

the  magnetizing  force  remains  the  same  as  before, 
_  (Bi  _  $1  _  0.47m/ 
\JL       AH  I 


FIG.  53.— Ring 
solenoid. 


FIG.  54.— Solenoid 


FIG.  55. — Solenoid  wound 
on  an  iron  bar. 


2.  Fig.^54,  represents  a  solenoid  of  n  turns  carrying  a  current  of  /  amp.; 
the  m.m.f.  is  M  =  0.47m/  gilberts.  If  I  cm.  is  the  length  of  the  solenoid 
and  A  sq.  cm.  is  its  sectional  area,  the  reluctance  of  the  path  through  it  is 


The  m.m.f.  required  to  drive  the  flux  <£  through  this  reluctance  is 

MI  =  <t>-r  gilberts. 
A. 

But  the  flux  going  out  at  one  end  has  to  pass,  around  through  the  air  and 
in  at  the  other  end  as  shown.  The  reluctance  of  this  return  path  is  difficult 
to  calculate  and  is  not  of  great  practical  importance;  its  length  is  greater 
than  I  but  its  sectional  area  is  very  much  greater  than  A  and  its  reluctance 
is  small  compared  to  (Ri.  Thus  in  the  case  of  long  solenoids  of  small  section 
the  reluctance  of  the  return  path  may  be  neglected  and  the  assumption  may 
be  made  that  the  whole  m.m.f.  M  is  utilized  in  driving  the  flux  through  the 
reluctance  (Ri.  Thus 


* 


MAGNETISM  AND  ELECTROMAGNETICS         63 


the  induction  density  in  the  solenoid  is 


0.47m/ 


and  the  magnetizing  force  at  any  point  inside  the  solenoid  is 
3C  =  (B  =    ' ,  ft*    dynes,  as  in  equation  (108). 

If  an  iron  bar  of  length  I  and  permeability  /*  is  placed  in  the  solenoid  the 
reluctance  of  the  path  through  the  solenoid  is  reduced  in  the  ratio  -  and  the 

reluctance  of  the  return  path  is  no  longer  negligible  in  comparison  with  it  and 
its  value  must  be  calculated  (Fig.  55). 


A.SQ.OI 


FIG.  56. — Ring  with  an 
air  gap. 


FIG.  57. — Magnetic  circuit 
of  a  dynamo. 


3.  In  Fig.  56  a  solenoid  of  nl  ampere-turns  is  wound  on  an  iron  ring  of 
permeability  IJL  and  a  section  of  length  12  is  cut  from  the  iron.  If  l\  is  the 
mean  length  of  the  path  through  the  iron  and  its  sectional  area  is  A  sq. 
cm.  the  reluctance  of  the  path  through  the  iron  is 

h 

^•••4? 
the  reluctance  of  the  path  through  the  air  is 


and  the  total  reluctance  of  the  path  is 
(R  =  (Ri  +  <R2 

The  flux  through  the  path  is 

i'mM.m 

(R 


A 


. 
A/i      A 


(109) 


The  flux  density  in  the  iron  is  (B  =  -j  and  is  the  same  as  in  the  air. 
The  magnetizing  force  or  m.m.f.  per  centimeter  in  the  iron  is 


and  in  the  air  it  is 


64  ELECTRICAL  ENGINEERING 

The  m.m.f.  consumed  in  the  iron  is 

Mi  =  »x  =  -p-  h 

AM 

and  the  m.m.f.  consumed  in  the  air  is 

3> 
M2  =  «2  =  j  lt. 

4.  Fig.  53  represents  the  magnetic  circuit  of  a  bipolar  dynamo.     It  con- 
sists of  a  number  of  parts  of  different  materials  as  follows : 
One  yoke  y  of  section  Ay,  length  lv,  and  permeability  MJ/, 
Two  pole  cores  c,c  of  section  Ac,  length  lc,  and  permeability  MC, 
Two  pole  pieces  p,p  of  section  Ap,  length  lp,  and  permeability  MP, 
Two  air  gaps  g,g  of  section  Ag,  length  10,  and  permeability  na  =  1, 
One  armature  a  of  section  Aa,  length  la,  and  permeability  MO- 
The  reluctance  of  these  parts  are,  respectively, 

ZO7                                           O7                                         o;                                           7 
,.,                    y      ,        (r,        _     *LC            ,„                "lp             m             Al>g          m                 l>a 
(Jiy    =•  —. >        Otc    —  ~A '        Olp    =   ~7 '        W.g    =  ~r>        Cna    =   1 ; 

AyfJLy  ACfJLC  ApfJLp  Ag  AafJLa 

and  the  reluctance  of  the  whole  circuit  is 

tK    ^—    (jiy    "i     (j\c      i~  Orp      \~  (Jig      \~  CHa« 

The  m.m.f.  M  is  provided  by  field  coils  placed  on  the  pole  cores  as  shown, 
and  the  flux  through  the  circuit  is 

=  M  =  M 


01         Oltf  +  (Rc  +  (Rp  +  (Rff  +  (R0 

and  is  equal  to  the  m.m.f.  divided  by  the  total  reluctance;  and  the  m.m.f. 
is 


=     My    +    Mc    +    Mp    +    Mg    +    Ma, 

where  Mv  is  the  part  of  the  total  field  m.m.f.  required  to  drive  the  flux 
through  the  yoke,  etc. 

The  m.m.f.  Mg  required  to  drive  the  flux  across  the  air  gaps  is  sometimes 
as  much  as  80  per  cent,  of  the  total  m.m.f. 

5.  Determine  the  reluctance  of  the  ring  in  Fig.  58  made  up  of  three  parts 
of  lengths  li,  h  and  h  cm.  respectively  and  sectional  areas  Ai,  At  and  A3 
sq.  cm.  and  permeabilities  MI>  M2  and  MS.  The  m.m.f.  of  the  solenoid  is  M. 

The  reluctance  of  section  (1)  is  (Ri  =  -^ — ; 

lz 
the  reluctance  of  section  (2)  is  (R2    =  —A — ; 


the  reluctance  of  section  (3)  is  (R3    =  ^ — ', 

the  flux  through  section  (1)  is  $1     =  /o~J 

M 
the  flux  through  section  (2)  is  *2  =  ^-; 

the  flux  through  section  (3)  is  <f>3     =  ^r°f 


MAGNETISM  AND  ELECTROMAGNETICS 

the  total  flux  through  the  ring  is 


=  M       M_       M_ 
(Hi        (R2        (Ra 


65 


j_     j_\ 

i       (R2       (R3/ 


(110) 


But  the  flux  is  equal  to  the  m.m.f.  divided  by  the  reluctance  of  the  ring,  or 

M 


and  therefore 


M 


(HI) 


FIG.  58. — Parallel  magnetic  paths.      FIG.  59. — Series-parallel  magnetic  circuit. 

6.    Fig.  59  shows  a  ring  of  iron  with  a  piece  set  in  made  up  of  three 
parts  of  different  permeabilities.     The  lengths,  sections  and  permeabilities 

are   indicated   and    the   reluctances   of  the  various  parts  are  (Ri  =  -r^— » 

«  AlMl 

and  (R4  =  -7  —  •     Determine  the  reluctance  of  the 


circuit. 

Let  M    —  the  m.m.f.  applied  to  the  ring, 
$     =  flux  through  the  ring, 
MI  =  m.m.f.  consumed  in  section  (1), 
M2  =  m.m.f.  consumed  in  section  ab, 
then     MI  =  3>(Ri,  and  M2  =  €>(R0b. 

The  reluctance  of  the  section  ab  consisting  of  three  paths  in  multiple  must 
be  found.  The  flux  $  passing  through  it  divides  into  three  parts  which  are 
inversely  proportional  to  the  reluctances  of  the  paths, 


therefore, 


and 


(Rs'         4        (R4' 


(Rab    = 


(R2       (R3       (R4 


66  ELECTRICAL  ENGINEERING 

The  reluctance  of  the  whole  circuit  is 

M    -  Af i  +  M 


<R2       (R3       (R4 

69.  Energy  Stored  in  the  Magnetic  Field. — When  a  current 
i  c.g.s.  units  flows  in  the  solenoid,  Fig.  60,  wound  on  an  iron  ring, 
a  flux  is  produced, 

4-nni  . 
<P  =  — 5 — Au, 

I 

where     n  is  the  number  of  turns  in  the  solenoid, 
A  is  the  sectional  area  of  the  ring, 
I  is  the  mean  length  of  the  ring  and 
fj.  is  its  permeability  and  is  treated  as  though  it  were  a 
constant. 


FIG.  60. — Solenoid  on  an  iron  ring. 

When  the  current  increases  by  a  small  amount  di  the  flux  in- 
creases by  an  amount  d<j>  =  — j —  tfoV.and  the  work  done  is  equal 

to  the  product  of  the  current  and  the  increase  of  the  flux, 

,             .  7          Aiirn^Au,  .  7. 
aw  =  ni  a<j>  = = 1  di  ergs. 

The  work  done  while  the  current  is  building  up  to  its  full  value 

/is 

i 


,-^ON 
-g  ergs  (112) 

(113) 

, 
where  £  = j is  a  constant,  if  //  is  constant,  and  is  called  the 

inductance  of  the  circuit. 


MAGNETISM  AND  ELECTROMAGNETICS         67 

This  amount  of  energy  is  stored  in  the  magnetic  field  of  the 
solenoid  and  may  be  expressed  in  other  forms. 
The  energy  stored  is 


TF  =  ^prJerSs' 
but  the  magnetic  force  in  the  solenoid  is 

5C  --j-' 

therefore, 

or  since  the  induction  density  is 

(B  =  jitfC, 
the  energy  may  be  expressed  as 

W  -  ®2   A] 
-  Q      Yit. 

%  07TJU 

The  product  Al  represents  the  volume  of  the  magnetic  field 
and  therefore  the  energy  stored  in  the  field  per  unit  volume  is 

W  ,,TP2 

_   /Zuv 

or 


or 

5m 

STT 


ergs.  (116) 


When  the  magnetic  field  is  produced  in  air  the  field  intensity 
3C  is  equal  to  the  flux  density  (B,  and  the  energy  stored  per  cubic 
centimeter  is 

3C2 
w  =  -g^ergs,  (117) 

or 

(B2 
w  =        ergs.  (118) 


70.  Stress  in  the  Magnetic  Field.  —  The  force  between  two 
magnetic  poles  is  not  exerted  at  a  distance  but  is  transmitted 
through  the  medium  separating  them  and  the  medium  is  stressed. 


68 


ELECTRICAL  ENGINEERING 


There  is  a  tension  along  the  lines  of  force  or  induction  tending 
to  shorten  them  and  draw  the  boundary  surfaces  of  the  field 
together  and  so  decrease  the  magnetic  energy  stored  in  the  field; 
there  is  also  a  pressure  at  right  angles  to  the  lines  tending  to 
spread  them  apart,  and  so  decrease  the  flux  density  in  the  field 
and,  therefore,  also  the  energy  stored.  Since  the  medium  is  in 
equilibrium  these  two  forces  are  equal  in  magnitude. 

The  pull  on  the  bounding  surfaces  is  usually  expressed  in  dynes 
per  square  centimeter  and  may  be  found  as  follows: 

The  energy  stored  in  a  magnetic  field  in  air  was  found  to  be 

(&2 

—  ergs  in  Art.  69 

07T 

Fig.  61  represents  a  horseshoe  magnet  with  its  armature  re- 
moved a  distance  x.  If  (B  is  the  flux 
density  in  the  field  between  the  magnet 
poles  and  the  armature  and  A  sq.  cm.  is 
the  area  of  the  two  pole  faces,  the  volume 
of  the  field  is  Ax  c.c.  and  the  energy 
stored  in  it  is  » 

W  =  •$-  Ax  ergs. 

O7T 

If  a  pull  P  is  exerted  on  the  armature 
and  it  is  moved  through  a  distance  dx,  the 
work  done  is  P  dx  and  this  is  equal  to  the 
increase  in  the  energy  stored  in  the  field; 
therefore, 


FIG.  61.— Pull  of  a 
magnet. 


Pdx  =    - 


and 


A 

dynes; 


this  is  the  pull  of  the  magnet  on  the  armature. 
The  pull  per  square  centimeter  is 

P       6 

P    =   -A     =    Q 


(119) 


(120) 


and  this  is  the  tension  along  the  lines  of  induction  in  the  field  and 
is  also  equal  in  magnitude  to  the  pressure  at  right  angles  to  the 
lines. 

71.  Force  between  Parallel  Wires  Carrying  Current. — Parallel 
wires  carrying  currents  /i  and  /2  absamp.  will  either  be  attracted 


MAGNETISM  AND  ELECTROMAGNETICS 


69 


or  repelled  depending  on  whether  the  currents  are  in  the  same  or 
opposite  directions.  Fig.  62  shows  the  fields  in  the  two  cases. 
When  the  currents  are  in  the  same  direction  the  lines  of  force 
combine  to  form  lines  surrounding  the  two  wires  and  these  lines 
tend  to  shorten,  or  the  circuit  tends  to  change  so  that  the  reluc- 
tance is  a  minimum  and  the  flux  a  maximum.  When  the  currents 
are  in  opposite  directions  the  lines  of  force  pass  between  the  wires 
and  they  tend  to  spread  out  resulting  in  a  repulsion  between  the 
wires.  In  this  case  again  the  forces  tend  to  decrease  the  reluc- 
tance of  the  circuit  by  increasing  its  section. 


p  -  d- 

FIG.  62.  —  Force  between  two  parallel  wires  carrying  current. 

To  obtain  an  expression  for  the  force  between  the  two  wires 
refer  to  Fig.  62.  A  and  B  are  the  two  wires  carrying  currents  /i 
and  /2  respectively  and  d  cm.  is  the  distance  between  their  centers. 

The  field  intensity  and  flux  density  at  wire  B  produced  by  the 
current  /i  in  A  is 

27 

(Bi  =~~T~  lines  per  square  centimeter. 

This  field  acts  on  the  wire  B  carrying  current  1%  witha  force 


f  =  (81/2  = 


O7"   T 


dynes  per  centimeter  length.        (121) 


This  is  the  force  of  repulsion  between  the  two  wires. 

72.  Magnetic  Characteristics.  —  If  an  iron  ring,  Fig.  60,  which 
has  been  completely  demagnetized,  is  gradually  magnetized,  by 
increasing  the  current  in  the  exciting  coil  from  zero,  the  induction 
density  in  the  ring  increases  with  the  magnetizing  force  as  shown 
in  curve  1,  Fig.  63.  With  feeble  magnetizing  forces  the  gradient 
of  the  curve  is  small  and  the  permeability  is  low;  as  the  force 
increases  the  curve  becomes  very  steep  and  nearly  straight  and 
the  permeability  increases  rapidly  and  becomes  very  large;  as 
the  force  is  further  increased  the  curve  rounds  off  and  the  gradi- 
ent becomes  small  again  and  a  large  increase  in  magnetizing 
force  is  required  to  produce  any  considerable  increase  in  density, 
the  permeability  decreases  to  a  low  value  and  the  material  is 


70 


ELECTRICAL  ENGINEERING 


said  to  be  saturated.     Curve  1  is  called  a  "magnetization  curve" 
or  " saturation  curve,"  or  (B-3C  curve  of  the  material. 

Curve  2,  Fig.  63,  shows  the  relation  between  the  permeability 

/T> 

M  =  —  and  the  induction  density  (B;  when  the  density  is  low 


Magnetizing  Force  =  3C 

FIG.  63. — Magnetic  characteristics. 


(Bi 

Magnetic  Induction  =  (B 


the  permeability  is  low;  as  the  density  is  increased  the  per- 
meability increases  until  it  reaches  a  maximum  at  the  point  m 
where  the  tangent  from  the  origin  touches  curve  1;  above  this 
point  the  permeability  decreases  again. 

In  Figs.  64  and  65  are  shown  magnetization  curves  for  materials 
used  in  electrical  machine  design.     These  curves  are  plotted  with 


100   200 


Permeability  -»  ft. 
400   500   600   700   800   900   1000  1100  1200  1300 


in 


100       120       140       160        180       200       220       240      2GO 
Ampere  Turns  per  Inch. 

FIG.  64. — Magnetic  characteristics  of  cast  iron. 

induction  density  B  expressed  in  lines  per  square  inch  on  a  base 
of  ampere-turns  per  inch  instead  of  magnetizing  force  JC.  This 
results  in  a  change  of  soales  only  since  B  lines  per  square  inch 
=  (2.54) 2  (B,  and  T  ampere-turns  per  inch  corresponds  to  a 

magnetizing  force  5C  =     '  ..  • 


MAGNETISM  AND  ELECTROMAGNETICS 


71 


Permeability  curves  for  cast  iron  and  cast  steel  are  also  shown. 
For  cast  iron  the  maximum  value  of  n  occurs  at  a  density  of  about 
4,000  lines  per  square  centimeter  or  25,000  lines  per  square  inch 
and  for  steel  at  about  6,500  lines  per  square  centimeter  or  40,000 
lines  per  square  inch.  Densities  below  these  points  are  not  of 
very  great  importance. 


Permeability  *= 

CO       0        100       200      300      400       500       600       700       800       900      1000      1100     1200  1300 
H  140 


400 
Ampere  Turns  per  Inch 

FIG.  65. — Magnetic  characteristics  of  cast  steel. 


600      650 


73.  Hysteresis. — If  the  exciting  current  i  or  magnetizing  force 
5C,  of  the  solenoid  in  Art.  69,  is  increased  the  flux  density  (B  in- 
creases, following  the  curve  OA  in  Fig.  66  until  it  reaches  its 
maximum  value  (BTO  at  A.  If  3C  is  now  gradually  reduced  to 
zero,  (B  will  not  come  back  on  AO  but  will  follow  the  curve  AR 
and  when  3C  is  zero  (B  will  have  the  value  (Br  represented  by  OR. 
This  is  called  the  residual  magnetism;  it  is  not  a  fixed  quantity 
but  depends  on  the  degree  of  magnetization  of  the  material. 
To  remove  the  residual  magnetism  5C  is  reversed  and  increased 
to  the  point  C.  (B  is  now  zero.  The  magnetizing  force  repre- 
sented by  OC  is  called  the  coercive  force.  As  5C  is  still  increased 
(B  increases  to  its  maximum  value  (Bm  again  at  the  point  A'  with 
the  same  magnetizing  force  as  at  A.  Again  reducing  3C,  (B  follows 
the  curve  AiRiCi  to  A  and  closes  the  loop.  This  closed  curve 
is 'called  a  hysteresis  loop. 

The  area  of  the  hysteresis  loop  represents  the  amount  of 
energy  consumed  in  carrying  the  material  through  the  cycle  of 
magnetism. 


72 


ELECTRICAL  ENGINEERING 


From  Art.  64  the  work  done  in  increasing  the  flux  threading 
the  solenoid  by  an  amount  d<$>  is 

,          ni  , 
dw  =  —  d<}>  ergs, 

Where  n  is  the  number  of  turns  on  the  coil  and  i  is  the  current 
in  amperes  when  the  flux  threading  the  coil  is  0.  If  the  ring 
has  a  constant  section  of  A  sq.  cm.  and  a  length  of  I  cm.,  then, 

4'7r??'Z'  "*^* 

since  3C  =  -^r  and  d$  X  Ad(&,  the  work  done  is 


10Z 


Al 
dw  =     - 


FIG.  66. — Hysteresis  loop. 

and  the  work  done  during  a  complete  cycle  is 
W  = 


JAl    C®m 
dw  =  ^p   I      3Cd(B  ergs. 
4?T    Jffim 


AZis  the  volume  of  the  ring  in  cubic  centimeters  and  therefore 
the  loss  in  ergs  per  cycle  per  cubic  centimeter  of  iron  is 


i 

Wh    =   -r- 


(122) 


Referring  to  Fig.  66  and  starting  from  the  point  Ci,  the  energy 
per  cubic  centimeter  supplied  to  increase  the  induction  density 


MAGNETISM  AND  ELECTROMAGNETICS          73 

from  0  to  (BTO  is  -r—  (area  OCiAM)  ergs.     While  (B  decreases  from 
(BTO  to  (Br,  energy  is  given  back  =  -r—  (area  AMR).     The  energy 

4r7T 

required  to  remove  the  residual  magnetism  is  ^—  (area  ROC). 
Similarly  below  the  axis  the  energy  consumed  is 

j—  (area  OCAiMi  —  area  AiMiRi  +  area  ORiCi)  ergs. 

Thus  the  hysteresis  loss  per  cubic  centimeter  per  cycle  is  j—  (area 

of  loop)  ergs. 

The  area  of  the  loop  increases  faster  than  the  maximum  den- 
sity. Steinmetz  gives  the  following  equation  for  the  loss  per 
cubic  centimeter  per  cycle  in  terms  of  the  maximum  density: 

COA  =  liffim^ergs,  (123) 

where  rj  is  called  'the  hysteretic  constant  of  the  material. 

The  hysteresis  loss  in  a  volume  of  V  c.c.  of  iron  at  a  frequency 
of  /  cycles  per  second  is 

Wh  =  i?(Bn1-8/Vr10-7  watts.  (124) 

The  following  table  gives  ordinary  values  for  the  hysteretic 
constant  for  commercial  materials. 

Good  dynamo  sheet  steel 0 . 002 

Fair 0.003 

Silicon  steel 0.00076-0.004 

Cast  iron 0.011-0.016 

Cast  iron 0.011-0.016 

Cast  steel 0.003-0.012 

Good  commercial  iron  after  assembly 0 . 0027 

Iron  can  be  carried  through  a  cycle  of  magnetism  or  hysteresis 
cycle  (1)  by  applying  an  alternating  m.m.f.  and  producing  an 
alternating  flux  as  in  the  transformer;  (2)  by  producing  a  revolv- 
ing flux  which  cuts  stationary  iron  as  in  the  stator  of  an  induction 
motor;  and  (3)  by  revolving  iron  in  a  stationary  magnetic  field 
as  in  direct-current  generators  and  motors. 

The  energy  expended  in  hysteresis  heats  the  iron  and  causes 
a  loss  in  efficiency. 

74.  Magnetic  Materials. — The  most  important  magnetic  mate- 
rials used  in  electrical  design  are  cast  iron,  cast  steel,  ordinary 
sheet  steel  and  silicon  sheet  steel. 


74  ELECTRICAL  ENGINEERING 

Cast  iron  has  low  permeability  and  large  hysteresis  loss.  It 
is  used  for  parts  of  the  magnetic  circuit  where  the  induction  den- 
sity is  low  and  is  in  a  constant  direction.  It  is  cheap  and  can  be 
made  into  castings  of  complex  form. 

Cast  steel  has  a  much  higher  permeability  and  smaller  hys- 
teresis loss.  It  replaces  cast  iron  where  greater  strength  or 
greater  permeability  is  required  or  when  the  appearance  would 
be  improved  by  reducing  the  section. 

Sheet  steel  is  used  for  parts  carrying  alternating  fluxes  of  all 
densities.  It  is  necessary  to  use  thin  sheets  in  such  places  to 
reduce  the  loss  due  to  eddy  currents  set  up  in  the  iron  as  it  cuts 
across  the  flux  (see  Art.  184).  To  further  reduce  the  eddy-cur- 
rent loss  the  iron  should  have  as  high  an  electrical  resistance  as 
possible. 

Silicon  steel  contains  a  small  percentage  of  silicon  which  has 
the  effect  of  reducing  the  hysteresis  loss  and  increasing  the  elec- 
trical resistance  and  so  reducing  the  eddy-current  loss.  Much 
higher  flux  densities  can  thus  be  used  and  the  weight  and  cost  of 
machines  reduced.  Silicon  steel  is  used  principally  for  trans- 
formers, where  very  high  efficiency  is  required.  It  is  quite 
expensive. 

75.  Effect  of  Chemical  Composition  and  Physical  Treatment 
on  Hysteresis  Loss. — Hysteresis  loss  is  dependent  on: 

1.  Chemical  composition  of  the  iron. 

2.  Heat  treatment. 

3.  Mechanical  treatment. 

1.  The  general  effect  of  impurities  is  to  decrease  the  permea- 
bility and  increase  the  iron  loss  but  there  are  exceptions. 

Carbon  decreases  the  permeability  and  lowers  the  saturation 
point;  it  increases  the  residual  magnetism  and  the  coercive  force. 
These  effects  are  greater  in  hardened  steel  than  in  soft  iron. 
Such  materials  are  valuable  for  permanent  magnets  but  are  not 
used  in  rotating  machinery. 

Silicon  in  certain  percentages  has  a  very  beneficial  effect;  2.5 
to  4  per  cent,  of  silicon  alloyed  with  the  iron  increases  the  per- 
meability and  decreases  the  hysteresis  loss  to  a  marked  degree. 
It  also  increases  the  electrical  resistance. 

T.  D.  Yensenhas  obtained  samples  of  iron,  alloyed  with  certain 
percentages  of  silicon  and  melted  in  vacuo,  which  exhibit  mag- 
netic qualities  far  superior  to  any  commercial  materials.  These 


MAGNETISM  AND  ELECTROMAGNETICS 


75 


materials  are  only  in  the  experimental  stage.     The  table  below 
shows  a  comparison  between  various  steels. 

Aluminum  has  an  effect  similar  to  that  of  silicon  but  the  im- 
provement is  not  so  great. 

2.  Annealing  the  material  after  manufacture  has  a  very  im- 
portant effect.     In  certain  cases  it  has  reduced  the  loss  in  sheet 
iron  to  half. 

It  is  found  that  the  hysteresis  loss  in  some  steels  increases 
with  continued  heating  during  use.  This  is  known  as  aging  and 
may  have  very  serious  results. 

Silicon  steel  is  non-aging. 

3.  Punching,  hammering  and  bending  increase  the  hysteresis 
loss  in  sheet  steel. 

The  following  table  shows  a  comparison  between  Yensen's 
experimental  steels  and  some  commercial  materials. 


Material 

Maximum 
permeability 

Density  for 
maximum 
permeability 
in  lines 
per  square 
centimeter 

Hysteresis  loss  in 
ergs  per  cubic 
centimeter  per 
cycle 

Electrical 
resistance 
in 
microhms 
per 
centimeter 
cube 

/o 
^max.  " 
10,000 

«max.= 
15,000 

Experimental  steel  with  0.15  per 
cent,  silicon  

66,500 

63,300 
3,850 
3,160 

6,500 

6,500 
7,000 
4,000 

286 

280 
3,320 
2,260 

916 

1,025 
5,910 
3,030 

11.8 

48.5 
11.0 
51.0 

Experimental  steel  with  3.40  per 
cent,  silicon  
Standard  transformer  steel  
4  per  cent,  silicon  steel  

76.  Theories  of  Magnetism. — The  magnetization  of  iron  is 
accompanied  by  molecular  changes  and  energy  is  consumed. 
Ewing  was  the  first  to  formulate  a  theory  to  explain  the  resulting 
phenomena.  He  imagined  iron  to  be  made  up  of  magnetic 
molecules.  In  unmagnetized  material  these  molecules  are  ar- 
ranged in  groups,  in  which  the  elementary  magnetic  poles  neu- 
tralize one  another.  When  a  weak  magnetizing  force  or  m.m.f .  is 
applied,  the  less  stable  groups  are  broken  up  and  their  component 
molecules  turned  with  their  magnetic  axes  in  the  direction  of  the 
impressed  m.m.f.,  making  the  material  as  a  whole  slightly  mag- 
netic. This  stage  is  represented  by  the  lowest  section,  A,  of  the 
magnetization  curve,  Fig.  63.  If  the  magnetizing  force  is  re- 
moved, the  molecules  return  to  their  original  groupings  and  no 
magnetism  remains.  When,  however,  an  increasingly  powerful 


76  ELECTRICAL  ENGINEERING 

m.m.f.  is  applied,  the  more  stable  groups  break  up  and  a  large 
proportion  of  the  molecules  turn  in  the  direction  of  the  external 
field  and  form  new  stable  groupings  ;  the  (B-3C  curve  rises  sharply 
and  the  permeability  is  large.  This  is  section  B  of  the  curve. 
If  at  this  point  the  m.m.f.  is  reduced  to  zero,  only  a  part  of  the 
molecules  resume  their  old  groupings,  the  majority  remaining  in 
their  new  positions.  In  this  way  the  phenomenon  of  residual 
magnetism  is  accounted  for. 

With  an  ever  increasing  external  m.m.f.  more  and  more  of  the 
molecules  line  up  and  the  material  approaches  saturation. 
Beyond  this  point  any  increase  in  m.m.f.  only  produces  an  in- 
crease in  flux  density  comparable  with  that  produced  in  a  non- 
magnetic material  (section  C). 

The  later  electron  theory  of  magnetism  replaces  Ewing's 
magnetized  molecules  by  electromagnets.  It  is  assumed  that 
each  molecule  of  iron  is  the  seat  of  an  electric  current  or^whirl  of 
electrons  and  therefore  exerts  a  small  m.m.f.  An  external  m.m.f. 
tends  to  draw  these  into  line  with  itself  as  discussed  above  and 
thus  increases  the  resultant  m.m.f.  of  the  system  and  co'rise= 
quently  the  flux  and  induction  density.  With  this  assumption-  — 
the  conception  of  permeability  is  unnecessary  as  increasing 
permeability  is  replaced  by  increasing  m.m.f.  which  has  the  same 
effect. 

77.  Lifting  Magnets.  —  Design  a  direct-current  electromagnet 
capable  of  lifting  a  block  of  cast  iron  weighing  2,000  Ib. 

The  pull  of  a  magnet  is  given,  by  equation  (119), 

(B2A 
P  =  ^  dynes 

where  (B  is  the  flux  density  in  the  air  gap  in  lines  per  square 
centimeter  and  A  is  the  area  of  the  pole  surfaces  in  square 
centimeters. 

1  Ib.  =  444,000  dynes 
and  therefore 


p  =  ^  =  444,000  X  2,000  =  888  X  106  dynes. 

07T 

Assuming  an  air-gap  density  of  6,000  lines  per  'square  centi- 
meters (39,000  lines  per  square  inch)  the  required  pole  area  is 

888  X  106  X  Sir 

--  (6  000)  2         =  6  6  sq<  Cm*  =          Sq>  m' 

If  the  magnet  is  made  in  the  shape  shown  in  Fig.  67  (a)  the 


MAGNETISM  AND  ELECTROMAGNETICS 


77 


area  of  each  leg  is 


9S.5 


4$  sq.  in.     The  remaining  dimensions 


may  be  chosen  to  make  the  magnet  of  a  suitable  shape  but  space 
must  be  left  for  the  winding. 

The  number  of  ampere-turns  required  for  the  exciting  winding 
may  be  found  as  follows: 

Assume  that  the  contact  between  the  magnet  and  the  mass  to 
be  raised  is  not  perfect  and  that  an  air  gap  of  J^-in.  is  left.  The 
ampere-turns  required  for  the  gap  =  0.3132  X  39,500  X0.25 
X  2  =  6,170.  (Art.  223).  Length  of  the  path  through  the  steel 
magnet  =  36  in.;  the  ampere-turns  per  inch  required  for  a 
density  of  39,500  lines  is  17  (Fig.  65)  and  the  total  ampere- 
turns  =  17  X  32  =  544.  Length  of  the  path  in  the  cast  iron 


Sliding 


Non-Magnetic  Plates' 


FIG.  67. — Lifting  magnets. 

may  be  assumed  to  be  24  in.  and  the  section  60  sq.  in.;  the 


flux  density  in  the    cast   iron   is 


39,500_X  48 
60 


=  31,600;   the 


ampere-turns  per  inch  =  32  (Fig.  64)  and  the  total  ampere- 
turns  for  the  cast  iron  =  32  X  24  =  768. 

The  exciting  ampere-turns  for  the  magnet  =  6,170  +  544 
+  768  =  7,482. 

In  order  that  the  coil  may  not  become  overheated  it  is  neces- 
sary to  limit  the  power  loss  to  a  value  which  can  be  radiated  from 
the  external  cylindrical  surface  of  the  coil  without  too  great  a 
temperature  rise.  For  a  magnet  in  continuous  service  a  loss  of 
about  0.7  watts  can  be  taken  care  of  by  each  square  inch  of  the 
radiating  surface.  The  length  of  the  coil  is  12  in.  and  the  per- 
iphery is  44  in.  and  thus  the  radiating  surface  is  44  X  12  =  528 
sq.  in.  The  allowable  loss  is  therefore  528  X  0.7  =  370 
watts.  If  the  voltage  is  220  volts,  as  is  usually  the  case,  the  cur- 

370 
rent  is  /  =  ^^  =  1.7   amp. 


220 


and   the  resistance  of   the  coil  is 


=  -^=130  ohms. 


78  ELECTRICAL  ENGINEERING 

«,  ,         e  ,  m       ampere-turns       7,480 

The  number  of  turns  =  T  =  -    Current    -  =  -^-  =  4,400, 

the  length  of  the  mean  turn  is  36  in.  =  3  ft.  and  the  length  of 
wire  is  4,400  X  3  =  13,200  ft. 

The  resistance  of  a  wire  is  give  by  equation  (131). 

length  in  feet 


D 

JK    =    p 


section  in  circ.  mils 


The  specific  resistance  p  may  be  taken  as  12  for  a  temperature 
of  60°C.  and  substituting  in  the  equation  above  the  section  of 
the  conductor  in  circular  mils  may  be  obtained  : 

130  =  12  X  4 


circ.  mils 
and 

.,         12  X  13,200       1  oon 
circ.  mils  =  -  '     -  =  1,220. 


Referring  to  the  wire  table  on  page  85,  it  is  seen  that  No.  19 
wire  has  a  section  of  1,288  circ.  mils  and  it  would  therefore  be 
used. 

Lifting  magnets  are  usually  made  ring-shaped  as  shown  in 
Fig.  67(6)  and  a  large  factor  of  safety  is  required  on  account  of  the 
varied  nature  of  the  material  to  be  handled. 


CHAPTER  III 
ELECTRIC  CIRCUITS 

78.  Ohm's  Law. — When  an  e.m.f.  is  applied  to  the  terminals 
of  a  conductor,  a  current  is  produced  which  is  directly  propor- 
tional to  the  e.m.f.  and  is  inversely  proportional  to  the  resistance 
of  the  conductor; 

/  =  -5  amp.  ( 125) 

XL 

where  /  is  the  current  in  amperes, 

E  is  the  e.m.f.  in  volts, 
R  is  the  resistance  in  ohms. 

This  is  Ohm's  Law. 

A  conductor  has  a  resistance  of  one  ohm,  when  an  e.m.f.  of 
one  volt  is  required  to  drive  a  current  of  one  ampere  through  it. 

When,  therefore  a  current  /  flows  through  a  resistance  R, 
e.m.f.  is  consumed;  ^ 

E  =  IR  volts.  (126) 

79.  Joule's  Law. — Whenever  a  current  flows  through  a  re- 
sistance, electric  energy  is  transformed  into  heat  energy.     The 
power  or  the  rate  at  which  energy  is  transformed  in  the  circuit 
is  equal  to  the  product  of  the  current  and  the  e.m.f.  consumed  in 
driving  the  current  through  the  resistance  of  the  circuit. 

P  =  El  watts, 
but  E  =  IR  and,  therefore, 

P  =  PR  watts;  (127) 

thus,  the  power  consumed  in  the  circuit  is  equal  to  the  square  of 
the  current  multiplied  by  the  resistance.     This  is  Joule's  Law. 

The  power  consumed  in  the  resistance  of  circuits  represents  a 
loss  of  power  except  in  such  cases  as  the  incandescent  lamp,  where 
it  is  utilized  in  producing  light,  or  the  electric  heater,  where  the 
heat  developed  is  applied  to  a  useful  purpose. 

79 


80  ELECTRICAL  ENGINEERING 

80.  Heat  Units. — When  a  current  of  /  amp.  flows  through  a 
resistance  of  R  ohms  for  a  time  t  sec.  heat  is  developed 

H  =  PRt  watt-sec,  of  joules.  (128) 

The  practical  heat  units  are  the  British  thermal  unit  and  the 
calorie. 

The  British  thermal  unit,  B.t.u.,  is  the  heat  energy  required 
to  raise  1  Ib.  of  water  1°F. 

1  B.t.u.  =  1,055  watt-sec.  =  1.055  kw.-sec.          (129) 
The  calorie  is  the  heat  energy  required  to  raise  1  gram  of  water 

1  calorie  =  4.2  watt-sec.  =  0.0042  kw.-sec.  (130) 

81.  Examples. — 1.  If  electric  energy  costs  lf£  cts.  per  kilowatt-hour, 
what  will  it  cost  to  raise  the  temperature  of  100  gal.  of  water  60°F.? 

One  gallon  of  water  weighs  8.4  Ib. 

The  energy  required  to  raise  100  gal.  =  840  Ib.  of  water  60°F.  is 

840  X  1,055  X  60  watt-sec.  =  -  '         *       kw.-hr. 

and  the  cost  is 

840X1,055  X60  _  _ 

1,000  X  3,600 

2.  If  2  tons  (2,000  Ib.)  of  coal  are  required  per  month  to  heat  a  house,  what 
would  it  cost  to  supply  the  same  amount  of  heat  electrically  at  1^  cts.  per 
kilowatt-hour? 

Assume  1  Ib.  of  coal  to  give  10,000  B.t.u. 

Heat  energy  in  2  tons  of  coal  =  2  X  2,000  X  10,000  B.t.u. 

2  X  2,000  X  10,000  X  1.055 

The  equivalent  kilowatt-hours  =  - — 

3,600 

2  X  2,000  X  10,000  X  1.055 
and  the  cost  =  - —  -  X  0.015  =  $175.83. 

82.  Resistance. — The  resistance  of  a  conductor  varies  directly 
as  its  length  and  inversely  as  its  sectional  area;  it  also  depends  on 
the  material  of  which  the  conductor  is  made; 

#  =  Pj->  (131) 

where 

I  is  the  length  of  the  conductor, 

A  is  the  sectional  area, 

p  is  the  specific  resistance  or  resistivity  of  the  material. 


ELECTRIC  CIRCUITS  81 

The  specific  resistance  or  resistivity  may  be  expressed  in  ohms 
per  centimeter  cube  or  per  inch  cube,  which  is  the  volume  resis- 
tivity; or  in  ohms  per  meter-gram,  i.e.,  the  resistance  of  a  uniform 
round  wire  1  meter  long  weighing  1  gram,  this  is  called  the  mass 
resistivity. 

In  engineering  problems  wires  are  usually  specified  by  gage 
numbers,  their  lengths  are  given  in  feet  and  their  sectional  areas 
in  circular  mils.  A  circular  mil  is  the  area  of  a  circle  one  mil  or 
one-thousandth  of  an  inch  in  diameter.  The  specific  resistance 
is  then  the  resistance  of  a  wire  one  foot  long  and  one  circular 
mil  in  section. 

83.  Conductance. — The  reciprocal  of  the  resistance  of  a  con- 
ductor is  called  its  conductance  and  is  represented  by  the  letter 
G,  where 

o-  l- 

-R 

The  reciprocal  of  the  specific  resistance  or  resistivity  of  a  material 
is  called  its  conductivity  and  is  represented  by  the  Greek  letter 
7,  where 

1 

7  =  -  ' 
P 

Conductivity  should  be  expressed  in  per  cent,  of  the  conduc- 
tivity of  the  International  Annealed  Copper  Standard  taken  as 
100  per  cent.  This  standard  is  expressed  in  terms  of  mass  re- 
sistivity as  0.15328  ohms  (meter-gram),  at  the  standard  tempera- 
ture 20°C.  This  is  equivalent  to  10.371  ohms  per  circ.  mil  foot 
at  20°C. 

The  conductivity  of  metals  is  usually  decreased  by  the  presence 
of  impurities.  Drawing  or  other  cold  working  makes  metals 
harder,  stronger  and  slightly  more  dense  and  decreases  their  con- 
ductivity. The  conductivity  of  copper  wires  ranges  from  96  to  98 
per  cent.  Aluminum  wire  has  a  conductivity  of  60  or  61  per  cent. 

84.  Effect  of  Temperature  on  Resistance. — The  resistance  of 
all  pure  metals  increases  with  increase  of  temperature.     The 
variation  can  be  expressed  by  the  formula 

Rt  =  Ro(l  +  at),  (132) 

where  R0  is  the  resistance  at  a  chosen  standard  temperature,  Rt 
is  the  resistance  at  a  temperature  t°  higher  and  a  is  the  tem- 
perature coefficient  of  resistance.  It  is  the  increase  in  resist- 


82  ELECTRICAL  ENGINEERING 

ance  per  degree  rise  in  temperature  expressed  as  a  fraction  of  the 
resistance  at  the  standard  temperature.  If  the  Centigrade  scale 
is  used  and  RQ  is  the  resistance  at  0°C.,  the  value  of  a  for  copper 
of  100  per  cent,  conductivity  is  0.00427.  This  value  may  be  con- 
sidered as  constant  over  the  range  of  temperature  from  0°  to 
100°C.  With  the  Fahrenheit  scale  and  RQ  taken  as  the  resist- 
ance at  32°F.,  the  value  of  a  is  -^-—^ —  =  0.00237. 

l.o 

If  the  temperature  t°C.  is  taken  as  standard  the  formula  may 
be  written 

Rt  =  Rtj{l  +a(t-  O},  (133) 

where  a  is  not  the  same  as  before  but  is  smaller  because  the  in- 
crease of  resistance  is  expressed  as  a  fraction  of  the  resistance  at 
ti°C.  which  is  greater  than  the  resistance  at  0°C. 

When  ti  is  taken  as  the  standard  temperature  20°C.  the  value 
of  a  is 

Ro 


R0  (1  +  a0  X  20)        1_  +  2Q 

OiQ 


234<+20 

Similarly,  starting  from  this  standard  temperature  coefficient 
at  20°C.,  the  value  of  a  at  any  temperature  t  is  found  as 

RZQ  OizQ  R%Q  1 


at   =   0:20 


5  +  t  -  20 


0.00393 

This  result  is  correct  if  the  conductivity  is  100  per  cent.  When 
it  is  less  than  100  per  cent,  the  value  of  a  must  be  varied  propor- 
tionally. For  copper  of  95  per  cent,  conductivity 

in.-1         -~  (135) 


' 


0.95  X  0.00393 

If  the  resistance  at  ti°C.  is  known,  but  the  corresponding  value 
of  atl  is  not  known,  the  resistance  at  any  other  temperature  t 
may  be  found  from 

1  +  0.00427*    \         D     /  234.5  +  t\      , 

0.00427  J  =  fl»  (mri)'  for  10°  per 

cent,  conductivity     (136) 


and 


ELECTRIC  CIRCUITS  83 

1  +  0.004277«  \  234.5  + 


234.5 


„ 


where  7  is  the  conductivity  expressed  as  a  decimal  fraction. 

If  the  resistance  Rtl  at  a  known  temperature  ti  is  given  and  also 
the  resistance  Rt  at  an  unknown  temperature  t,  this  tempera- 
ture t  may  be  found  as  follows: 

Rt^       R0(l+  0.00427*  )  _  234.5  +  t 
Ru  " 


or 


and 


R0  (I  +  0.00427*i)        234.5  + 
Rt  —  Rti  t  —  ti 


Rt 


234.5 


t-t,  =  ^^  (234.5  +  «i). 
Kt 


If  the  conductivity  is  7, 


Rt  -  Rti  /234.5 


(138) 


(139) 


85.  Properties  of  Conductors. — The  properties  of  the  most  im- 
portant electrical  conductors  are  given  in  the  table  below. 


Material 

Temperature  coefficient 

Density,  grams 
per  (cm.)  3 

Resistance,    ohms   per 
(cm.)  3  at  0°C. 

Aluminum  
Cooper  . 

0.00390 
0.00393 
0.00387 
0.00381 
0.00072 
0.00367 
0.00377 
0.00500 

2.70 
8.89 
11.36 
1.69-1.75 
13.55 
21.20-21.70 
10.40-10.60 
19.30 

2.64 

1.59 
18.40-19.60 
4.10-5.00 
94.07 
9.00-15.50 
1.50-1.70 
5.50 

>  X  10-" 

Lead  

Magnesium  .  .  . 
Mercury  
Platinum  
Silver  .... 

Tungsten  

The  temperature  coefficients  of  all  pure  metals  are  positive  and 
are  quite  large.  Certain  alloys  such  as  manganin  have  very 
small  temperature  coefficients  and  are  used  in  the  design  of  elec- 
trical instruments  where  the  resistance  must  remain  constant 
over  the  ordinary  range  of  temperature. 

Carbon  unlike  the  metals  has  a  negative  temperature,  coeffi- 
cient and  it  is  not  constant. 

The  resistance  of  insulating  materials  decreases  very  rapidly 
with  increase  of  temperature,  but  the  variation  is  not  regular  and 


84 


ELECTRICAL  ENGINEERING 


cannot  be  expressed  by  a  simple  equation.     Fig.  68  shows  the 
variation  of  the  resistance  of  slot  insulation  with  temperature. 
The  resistance  is  expressed  in  megohms  or  millions  of  ohms. 
86.  Resistance  of  Conductors. — The  resistance  of  a  circular 

mil  foot  of  copper  of  100  per  cent. 

conductivity  at  20°C.  is  10.371  ohms; 

therefore,  the  resistance  of  a  copper 

wire  at  20°C.  is,  by  equation  (131), 

length  in  feet 


200 

180 

B160 

£140 


£  loo 
I  80 

S    60 


R  =  10.371 


=  10.371 


section  in  circular  mils. 


0    10    20    30   40    50  60  70    80  90  100 

Temperature  Degrees  Centigrade 

FIG.  68. — Variation  of  resist- 
ance of  slot  insulation  with 
temperature. 


circ.  mils. 

In  the  case  of  rectangular  conductors 
such  as  busbars,  the  section  is  ex- 
pressed in  square  mils  and  the  value 
of  p  is  then  the  resistance  of  a  square 
wire,  one  mil  on  each  side  and  one  foot  long. 

The  following  table  gives  ;the  specific  resistance  of  copper  at 
various  temperatures  and  the  corresponding  values  of  the  tem- 
perature coefficient. 


Temperature 
in  degrees  C. 

Resistance  of  one 
circular  mil  foot  in 
ohms 

Resistance  of  one 
square  mil  foot  in 
ohms 

Temperature 
coefficient  a 

0 

9.560 

7.52 

0.00427 

15 

10.160 

7.98 

0.00401 

20 

10.371 

8.13 

0.00393 

25 

10.550 

8.30 

0.00385 

50 

11.590 

9.11 

0.00351 

75 

12  .  620 

9.85 

0.00323 

In  this  table  and  in  wire  tables  generally  the  conductivity  is 
assumed  as  100  per  cent.  If  the  conductivity  of  a  given  wire  is 
less  than  this  the  specific  resistance  must  be  changed  accordingly; 
thus,  the  resistance  of  one  circular  mil  foot  of  copper  of  98  per 

cent,  conductivity  at  25°C.  is  ~  ''  =  10.76  ohms;  the  resistance 
of  one  circular  mil  foot  of  aluminium  of  60  per  cent,  conductivity 
at  25°C.  is  =  17.58  ohms. 


ELECTRIC  CIRCUITS 


85 


TABLE  OF  COPPER  WIRE  OF  100  PER  CENT.  CONDUCTIVITY 


Ameri- 
can Wire 
gage 
(B.  &  S.) 
No. 

Diameter 
in  mils 
at  20°C. 

Section  at  20°C. 

Resistance, 
ohms  per 
1,000  ft.  at 
20°C. 

Current  capacity, 
amperes 

Circular 
mils 

Square 
inches 

Rubber- 
covered 

Other 
insulation 

0000 

460.00 

211,600.0 

0.1662 

0.04901 

225 

325 

000 

409.60 

167,800.0 

0.1318 

0.06180 

175 

275 

00 

364.80 

133,100.0 

0.1045 

0.07793 

150 

225 

0 

324.90 

105,500.0 

0.08289 

,0.09827 

125 

200 

1 

289.30 

83,690.0 

0.06573 

0.1239 

100 

150 

2 

257.60 

66,370.0 

0.05213 

0.1563 

90 

125 

3 

229.40 

52,640.0 

0.04134 

0.1970 

80 

100 

4 

204.30 

41,740.0 

0.03278 

0.2485 

70 

90 

5 

181.90 

33,100.0 

0.02600 

0.3133 

55 

80 

6 

162.00 

28,250.0 

0.02062 

0.3951 

50 

70 

7 

144.30 

20,820.0 

0.01635 

0.4982 

8 

128.50 

16,510.0 

0.01297   - 

0.6282 

35 

50 

9 

114.40 

13,090.0 

0.01028 

0.7921 

10 

101.90 

10,380.0 

0.008155 

0.9989 

25 

30 

11 

90.74 

8,234.0 

0.006467 

1.260 

12 

80.81 

6,530.0 

0.005129 

1.588 

20 

25 

13 

71.96 

5,178.0 

0.004067 

2.003 

14 

64.08 

4,107.0 

0.003225 

2.525 

15 

20 

15 

57.07 

3,257.0 

0.002558 

3.184 

16 

50.82 

2,583.0 

0.002028 

4.016 

6 

10 

17 

45.26 

2,048.0 

0.001609 

5.064 

18 

40.30 

1,624.0 

0.001276 

6.385 

3 

5 

19 

35.89 

1,288.0 

0.001012 

8.051 

20 

31.96 

1,022.0 

0.0008023 

10.15 

21 

28.46 

810.1 

0.0006363 

12.80 

22 

25.35 

642.4 

0.0005046 

16.14 

23 

22.57 

509.5 

0.0004002 

20.36 

24 

20.10 

404.0 

0.0003173 

25.67 

25 

17.90 

320.4 

0.0002517 

32.37 

26 

15.94 

254.1 

0.0001996 

40.81 

27 

14.20 

201.5 

0.0001583 

51.47 

28 

12.64 

159.8 

0.0001255 

64.90 

29 

11.26 

126.7 

0.0000995 

81.83 

30 

10.03 

100.5 

0.0000789 

103.2 

86 


ELECTRICAL  ENGINEERING 


87.  Drop  of  Voltage  and  Loss  of  Power  in  a  Distributing  Circuit— 
The  distributing  circuit  in  Fig.  69  delivers  20  kw.  at  220  volts  to  a  re- 
ceiver circuit  1,000  ft.  distant;  if  the  size  of  the  conductors  is  No.  1 
B.  &  S.,  determine  the  voltage  required  at  the  generating  end  of  the 
line  and  the  power  lost  in  the  line. 

Eg  is  the  generator  voltage, 

E  is  the  receiver  voltage  =  220  volts, 

I    is  the  load  current, 

r    is  the  resistance  of  each  conductor. 


FIG.  69.  —  Distributing  circuit. 
The  power  delivered  is 


the  current  is  therefore 


El  =  20,000  watts; 


20,000 
=      220      =  90<9 

the  resistance  of  each  conductor  at  25°C.  is 

r  =  0.126  ohms; 
the  voltage  drop  in  each  conductor  is 

el  =  Ir  =  90.9  X  0.126  =  11.48  volts; 
the  generator  voltage  is  therefore 

E0  =  E  +  2Ir  =  220  +  22.96  =  242.96  volts; 
the  drop  of  voltage  in  the  circuit  is 
2ei  =  27r  =  22.96  volts 

0°  per  cent<  =  9-4^  per  cent' 


22  96 


The  loss  of  power  in  the  circuit  is 

2/2r  =  2  X  90.92  X  0.1261  =  2,100  watts; 
the  power  delivered  by  the  generator  is 

Egl  =  242.96  X  90.9  =  22,100  watts; 
therefore  the  power  loss*  is 

2  100 

22TOO  *  10^  per  Cent*  =  9'45  per  cen^' 

and  the  efficiency  of  the  transmission  is  90.55  per  cent. 


ELECTRIC  CIRCUITS  87 

If  the  drop  of  voltage  had  been  limited  to  10  volts  what  size  of  wire 
would  have  been  required? 
The  drop  in  voltage  is 

1  000 

2Ir  =  2  X  90.9  X  10.55  ^  -~r  =  10  volts: 
circ.  mils 

therefore  the  required  section  in  circular  mils  is 

1  000 
A  =  2  X  90.9  X  10.55  -^  =  192,000  circ.  mils. 

88.  Current-carrying  Capacity  of  Wires.  —  The  energy  con- 
sumed in  the  resistance  of  a  wire  raises  the  temperature  of  the 
wire  until  the  point  is  reached  where  the  heat  radiated  and  con- 
ducted from  the  wire  is  equal  to  the  heat  generated  in  it.     When 
the  wire  is  bare  the  heat  will  escape  easily  into  the  air,  but  when 
it  is  covered  with  insulating  material  the  heat  cannot  escape  so 
easily  and  for  a  given  current  density  the  temperature  rise  will 
be  greater.     This  increase  in  temperature  decreases  the  resist- 
ance of  the  insulating  material  and  so  decreases  its  insulating 
properties;  in  extreme  cases  the  insulation  may  be  charred  and 
rendered  useless.     The  last  two  columns  of  the  table  in  Art.  86 
give  the  values  of  current  which  can  be  carried  safely  by  differ- 
ent sizes  of  wire.     With  rubber  insulation  the  allowable  current 
is  about  25  per  cent,  less  than  with  weatherproof  insulation  be- 
cause the  rubber  is  more  easily  affected  by  heat.     When  the  in- 
sulated wires  are  inclosed  in  conduits  the  current-carrying  capac- 
ity is  less  than  that  given  in  the  table. 

89.  Examples.  —  1.  If  the  resistance  of  a  copper  conductor  at  25°C.  is 
10  ohms,  determine  its  resistance  at  65°C. 

fles  =  #25{l  +  «25(65  -  25)}  =  10{l  +  0.00385(65  -  25)}  = 

11.55  ohms. 

2.  Determine  the  resistance  of  a  copper  wire  of  97  per  cent,  conductivity, 
100,000  circ.  mils  in  section  and  50  ft.  in  length  at  50.°C. 

,    The  resistance  is  =  0.00597  ohms. 


3.  Determine  the  resistance  of  an  aluminum  bar  0.75  in.  by  0.375  in.  by 
100  ft.  at  25°C.,  if  the  conductivity  is  60  per  cent. 
The  resistance  is 

Q    O  100 


4.  If  the  resistance  of  the  shunt-field  winding  of  a  generator  is  30  ohms 
at  a  temperature  of  25°C.  and  after  running  under  load  is  found  to  be  31.5 
ohms,  determine  the  average  temperature  of  the  winding. 


88  ELECTRICAL  ENGINEERING 

If  /  is  the  average  temperature  of  the  winding  when  hot,  'equation  (133) 
gives 

Rt  =  #25(1  +  0.00385(f  -  25)}, 
or  substituting 

31.5  =  30{l  +  0.00385(<  -  25)}, 

the  rise  in  temperature  is 

31.5  _ 


0.00385 
and  the  average  temperature  of  the  winding  is 

t  =  25  +  13  =  38°C. 

This  method  is  used  in  measuring  the  temperature  rise  in  the  field  and 
armature  windings  of  electrical  machines. 

90.  Kirchoff's  Laws.  —  Two  laws  enunciated  by  Kirchoff  are  of 
great  value  in  solving  problems  dealing  with  continuous-current 
circuits. 

First  Law.  —  The  algebraic  sum  of  all  currents  flowing  toward 
or  away  from  any  junction  is  zero. 

Second  Law.  —  The  algebraic  sum  of  the  e.m.fs.  acting  in  a 
closed  circuit  is  equal  to  the  algebraic  sum  of  the  products  of 
the  current  and  resistance  in  the  several  parts  of  the  circuit. 


VV         H 

(«)  (b) 

FIG.  70. — Kirchoff's  laws. 

The  first  law  is  illustrated  by  the  circuit  ABCD  in  Fig.  70 (o) 
The  current  i  divides  at  B  into  two  parts  ii  and  i2.,  iis  numerically 
equal  to  the  sum  of  i\  and  iz  and  therefore  the  algebraic  sum  of  all 
currents  at  the  junction  B  is 

i  —  i\  —  i<2.  =  0. 

The  second  law  is  illustrated  by  the  circuit  FGHK  in  Fig. 
70(6).  Between  the  points  F  and  G  an  e.m.f.  E  is  applied  which 
drives  a  current  i  around  the  circuit  in  a  clockwise  direction 
against  the  e.m.f.  e  of  the  battery  between  H  and  K.  r\  and  r2 
are  the  resistances  of  the  circuit. 


ELECTRIC  CIRCUITS 


89 


The  e.m.f.  acting  in  the  direction  of  the  current  is  E  —  e  and 
is  consumed  in  driving  the  current  i  through  the  resistances 
ri  and  rz  and  by  Ohm's  law 


E  —  e  = 


+  ir2. 


91.  Examples.  —  (A)  Fig.  71  shows  a  three-wire  system  supplied  by  two 
similar  110-volt  generators.  The  resistances  in  the  various  parts  of  the 
circuit  are  indicated.  Find  the  currents  in  the  three  lines,  assuming  the 
direction  of  flow  to  be  as  indicated  by  the  arrows. 

From  circuit  abgf,  110  =  ii(n  +  r3  +  RJ  +  i2r4or  110  =  8.3ii  +  0.4i2.(\) 
From  circuit  fgcd,  110  =  i3(r2  +  n  +  R2)  -  i2r4or  110  =  10.3i3  -  0.4i2.(2) 
From  circuit  abed,  220  =  ii(ri  +  r3  +  Ri)  +  is(Rz  +  r&  +  r2) 

or  220  =  8.3ii  +  10.3i3.  (3) 

At  the  point  g,  i\  =  i*  +  is  (4) 

Substituting  (4)  in  (1)       110  =  8.7i2  +  8.3i3  (5) 

Solving  (2)  and  (5),  i2  =  2.4  amp.,  i3  =  10.9  amp.  and  ii  =  i2  +  is  = 

13.3  amp. 
(B)  Find  the  currents  in  the  various  branches  of  the  circuit  in  Fig.  72. 


.**       r8=o.i 


,  C   . 


FIG.  71. 


FIG.  72. 


Equating  the  voltages  around  the  various  circuits  to  zero. 
(ABCDG)  2i    +  5ii  +  Mi  =  100. 

(ABFDG)  2i   +  4i2  +  5i5  =  100. 

(BCF)  5ii  +  3i3  -  4i2  =  0. 

(FCD)  3i4  -  5i5  -  3i3  =  0. 

(BCDF)  5ii  +    3i4  —  5i5  —  4i2  =  0. 

Equating  the  currents  at  the  junction  points  to  zero. 

(-B)  i    —  ii  —  i2  =  0,  or  i    =  i\  -f~  i2. 

f/~1\  •  •  •  n 

\ff/  *i  —  i3  —  ^4  =  u,  or  i\  =  ii  —  i3. 

(i))  i    —  i4  —  i5  =  0,  or  i    =  i4  +  i5. 

/P\  ...         .        ...         .         ... 

\"  /  ^2  T  *3  —  *6  ==  ",  or  15  =  i2  -f~  i3. 


(1)  * 

(2) 
(3) 
(4) 
(5) 


(6) 
(7) 
(8) 
(9) 


Substituting  (7)  and  (9)  in  (4), 
3  (ii  -is)   -  5  (i2  +  is)  -  3i3  =  0,  or  3ii  -  5i2  -  Ili3  =  0.  (10) 

Substituting  (7)  and  (9)  in  (5), 

5ii  +  3  (ii  -  is)  -  5  (ia  +  is)  -  4i2  =  0,  or  8ii  -  9i2  -  8i3  =  0.        (11) 


90  ELECTRICAL  ENGINEERING 

Solving  (3),  (10)  and  (11)  simultaneously, 

ii.  -  6.77  amp.,  i2  =  8.66  amp.  and  ia  =  —  1.76  amp. 
From  (6),  i  =  15.43;  from  (7),  u  =  8.53;  from  (9),  i*   =  6.90. 

92.  Resistances  in  Series.  —  If  a  voltage  E  is  applied  across 
a  circuit  consisting  of  a  number  of  resistances  Ri,  R2  and  Rs  con- 
nected in  series  (Fig.  73)  the  total  resistance  of  the  circuit  will 
be  equal  to  the  sum  of  the  resistances  of  the  different  parts. 

The  drop  of  voltage  across  the  resistance  Ri  is 


where  I  is  the  current  in  the  circuit. 
Similarly  ^  =  JR^ 

EZ  =  IRz, 

E    =   EI  +  Ez  + 


and  the  resistance  of  the  whole  circuit  is 

R*.  (141) 


FIG.  73. — Resistances  in  series.       FIG.  74. — Resistances  in  parallel. 

93.  Resistances    in   Parallel. — Determine    the    resistance    of 
t  a  circuit  consisting  of  a  number  of  resistances  Ri,  R%  and  Rs 
(Fig.  74)  connected  in  parallel. 

If  an  e.m.f  E  is  applied  to  the  circuit  a  current  /  will  flow 
which  wijl  divide  up  into  branch  currents  /i,  /2  and  73.  By 
Ohm's  law 

I    =  —  i 
I        —> 

'•-!• 

ZI3 

and  ^r 

where  E  is  the  resistance  of  the  whole  circuit, 


ELECTRIC  CIRCUITS 


91 


but 


therefore, 


/    =   /I   +  /2  +  /3 

\RI     RZ     R& 


R  = 


E 


\RI     RZ     RZ> 


- 

RI 


(142) 


If  the  conductances  G,  GI,  G2  and  G3  are  used  instead  of  the  terms 
ID>  ^D~>  D~>  and  ^ET  the  result  can  be  written 

K     £ll     £12  ^3 


R  =       - 


+  £2  +  Gs 


or 


(143) 


that  is,  the  conductance  of  a  circuit  consisting  of  a  number  of 
parallel  branches  is  equal  to  the  sum  of  the  conductances  of  the 
branches. 

94.  Potentiometer. — The  potentiometer 
shown  diagrammatically  in  Fig.  75  may 
be  used  to  obtain  a  variable  voltage  from 
a  constant  voltage  supply  in  a  continuous- 
current  system.  By  changing  the  position 
of  the  point  B  any  voltage  EAB  from  zero 
to  the  value  of  the  constant  supply  voltage 
E  may  be  applied  to  {he  load  resistance  RI. 

The  resistance  of  the  potentiometer  from 
A  to  B  is  R2  and  from  B  to  C  is  R5.  The 
various  currents  are  indicated  in  the  figure. 


FIG.  75. — Potentiometer. 


The  resistance  of  the  circuit  between  A  and  B  is 

1 


the  resistance  between  A  and  C  is 


R 


AC 


RAB  +  RBC  = 


i*i 

K\        K 


. 


92  ELECTRICAL  ENGINEERING 

and  the  current  from  the  mains  is 

T         E  E 

1  =  ~P~  =      P  p  ~      — J 
JtiAc        K\KZ 

the  drop  of  potenti?!  from  B  to  C  is 

j7i  TD     &K$ 

•&BC   —   ***J    —          r»    T-. 


K\K'z  „ 

^       i      r>     ~T  «-3 

and  the  drop  across  the  load  circuit  is 

77T  _      ~T7I          _      TjJ  _      Tjl 


,       ,-, 


There  is  a  continuous  waste  of  power  in  the  resistances  R2  and  #3  and 
the  efficiency  of  the  arrangement  is  always  less  than  the  ratio  of  the 
variable  voltage  EAB  to  the  constant  voltage  E. 

95.  Inductance.  —  When  a  current  of  electricity  flows  in  a  cir- 
cuit, lines  of  magnetic  induction  interlink  with  the  circuit;  if  the 
current  remains  constant  the  flux  threading  the  circuit  is  con- 
stant, but  when  the  current  varies  the  flux  varies  proportionally 
and  in  doing  so  generates  in  the  circuit  an  e.m.f.  proportional  to 
the  rate  of  change  of  the  interlinkages  of  flux  and  turns  or  to  the 
product  of  the  turns  and  the  rate  of  change  of  the  flux.  This 
e.m.f.  has  a  value 

6  =  —  n  -57  c.g.s.  units,  (144) 

where  n  is  the  number  of  turns  in  the  circuit  and  -57  is  the  rate 

of  change  of  the  flux  interlinking  with  the  turns.  The  negative 
sign  is  used  because  the  e.m.f.  produced  opposes  the  change  in 
the  current  and,  therefore,  the  change  in  the  flux. 

The  inductance  of  a  circuit  is  defined  as  the  rate  of  change  of 
the  flux  interlinkages  with  the  current  in  the  circuit  and  is  repre- 
sented by  <£,  thus, 

d<j> 

£  =  n-f-.  c.g.s.  units. 
ai 

This  quantity  is  not  a  constant  except  in  circuits  which  do  not 
contain  any  magnetic  materials  and  in  which  therefore  the  flux 
varies  directly  with  the  current. 

In  such  circuits  the  inductance  may  be  defined  as  the  number 


ELECTRIC  CIRCUITS  93 

of  interlinkages  of  flux  and  turns  for  unit  current  in  the  circuit, 
or, 

£  =  ^  c.g.s.  units,  (145) 

where         i  =  current  in  c.g.s.  units, 

<f>  =  flux  produced  by  current  i, 
n<f>  —  interlinkages  for  current  i. 

Since  £  =  -n  -jr  =  n-jr  /  -j->  the  e.m.f.  generated  in  a  circuit  due 
ai  at  /    at 

to  its  inductance  is 

e=  —  n-jT  =  —  £  -r  c.g.s.  units,  (146) 

It  is  equal  to  the  product  of  the  inductance  of  the  circuit  and  the 
rate  of  change  of  the  current.  When  e  is  expressed  in  volts  and 
i  in  amperes  the  inductance  is  in  henrys  and  is  represented  by  L 
to  distinguish  it  from  the  inductance  <£  expressed  in  c.g.s.  units. 
Equation  (146)  may  then  be  written 

e  =  -  L  jt  volts.  (147) 

The  henry  is  defined  as  the  inductance  of  a  circuit  in  which  a 
counter  e.m.f.  of  one  volt  is  generated  when  the  current  is  chang- 
ing at  the  rate  of  one  ampere  per  second. 

The  relation  between  the  two  units  of  inductance  is  found  as 
follows  : 

_  di 
e  =  —  £  -T  c.g.s.  units, 

=  —  <£  ^.10~8  volts,  £  and  i  in  c.g.s.  units, 

di 

=  —  £  ~r  10~9  volts,  £  in  c.g.s.  units,  i  in  amperes, 


and,  therefore, 

1  henry  =  109  c.g.s.  units  of  inductance.  (148) 

96.  Example.  —  (a)  Find  the  inductance  of  an  endless  solenoid  in  the  form 
of  a  ring,  Fig.  53. 

n  =  number  of  turns  in  solenoid  =   1,000. 
r  =  mean  radius  of  the  ring  =  10  cm. 

Z  =  27iT  =  mean  length  of  the  flux  path  through  the  solenoid. 
,4  =  sectional  area  of  the  solenoid  =  5  sq.  cm. 
i  =  current  in  solenoid  in  c.g.s.  units. 


94  ELECTRICAL  ENGINEERING 

The  flux  through  the  solenoid  is 

m.m.f.  4r7rru ,. 

^  =  T7T  lmes> 


reluctance        I/A 
the  flux  per  unit  current  is 


the  number  of  interlinkages  of  flux  and  turns  per  unit  current  is 

Z 

and  is  equal  to  the  inductance  of  the  circuit  in  c.g.s.  units;  thus,  the  induct- 
ance of  a  circuit  is  proportional  to  the  square  of  the  number  of  turns. 
Substituting  the  values  given  above  the  inductance  of  the  solenoid  is 

4  X  3.14  X  (1,000)2  X  5  = 
2  X  3.14  X  10 

and  the  inductance  in  practical  units  is. 

L  =  106  X  ID-9  =  0.001  henry. 

(6)  Find  the  energy  stored  in  the  magnetic  field  of  the  solenoid  when  the 
current  has  reached  a  value  /  =  10  c.g.s.  units. 

The  energy  stored  when  any  current  i  is  flowing  is  equal  to  the  work  done 
in  building  up  the  current  i  against  the  counter  e.m.f.  of  inductance  e  = 

—  £  -j-'  it  is  therefore 
dt 

W  =    \  eidt  =    I  £  ^  i  dt  =  £  \    i  di 
J  J      dt  Jo 


When  the  current  is  i  =  /  =  lOc.g.s.  units,  the  energy  stored  is 

/2  1Q2 

W  =  £  -2   =  106  X  -y  =  5  X  107  ergs 
=  5  X  107  X  10~7  =  5  watt-sec. 

97.  Inductance  of  Circuits  Containing  Iron. — As  stated  in  Art. 
95,  the  inductance  of  circuits  containing  magnetic  materials  is 
not  constant,  since  the  flux  does  not  vary  directly  with  the  cur- 
rent. The  relation  between  flux  and  current  may  be  represented 
by  a  magnetization  curve  or  saturation  curve  such  as  those  shown 
in  Art.  72,  but  these  curves  cannot  be  represented  by  mathemat- 
ical equations  and  for  engineering  calculations  it  is  necessary  to 
choose  some  constant  average  value  to  represent  the  inductance 
01  a  circuit,  that  is,  to  assume  that  the  material  has  a  constant 
permeability  M  and  then  to  apply  a  correction  to  the  result. 

If  a  solenoid  of  n  turns  is  wound  on  an  iron  ring  of  section  A 


ELECTRIC  CIRCUITS  95 

sq.  cm.  and  mean  length  I  cm.  and  permeability  /*,  the  flux  pro- 
duced by  a  current  i  c.g.s.  units  is 


,. 
lines, 

Ait 

the  flux  per  unit  current  is 


., 

c.g.s.  units, 
& 

and  the  inductance  is 


£  =  n<j>i  =  -  —  =  --  c.g.s.  units. 

It  is  proportional  to  the  square  of  the  number  of  turns,  as  seen 
in  the  last  example,  but  it  also  depends  on  the  permeability  of  the 
material  and  is  therefore  not  a  constant  quantity. 

When  the  current  in  the  solenoid  is  small,  the  iron  is  unsatur- 
ated,  the  permeability  is  high,  the  flux  per  unit  current  is  large 
and  therefore  the  inductance  is  large.  When  the  current  is  large, 
the  iron  is  saturated,  the  permeability  is  low,  the  flux  per  unit 
current  is  small  and  therefore  the  inductance  is  small. 

Thus  the  inductance  of  circuits  containing  iron  or  other  mag- 
netic materials  is  not  constant  but,  above  a  certain  point, 
decreases  as  the  current  increases  and  the  induction  density 
increases. 

The  permeability  curves  in  Figs.  64  and  65 
show  the  relation  between  permeability  and 
induction  density  for  cast  iron  and  cast  steel. 

98.  Mutual  Inductance  and  Self-induc- 
tance. —  If  two  electric  circuits  as  A  and  B  in 
Fig.  76  are  interlinked  with  a  magnetic  circuit 
of  constant  permeability  /*,  or  constant  mag- 
netic reluctance  (R,  the  mutual  inductance  of 
one  circuit  with  respect  to  the  second  circuit 
may  be  defined  as  the  number  of  interlinkages  of  the  second 
circuit  with  the  flux  produced  by  unit  current  in  the  first  circuit. 
It  is  equal  to  the  mutual  inductance  of  the  second  circuit  with 
respect  to  the  first  and  is  represented  by  3TC. 

The  flux  produced  by  unit  current  in  the  n\  turns  of  the  circuit 
A  is 


96  ELECTRICAL  ENGINEERING 

and  the  inductance  of  A  is 

£1  =  ni4>i  =  -  — ^-  c.g.s.  units;  (149) 

ot 

if  all  the  flux  produced  by  A  passes  through  B,  then  the  mutual 
inductance  of  A  with  respect  to  B  is 

3TlAs  =  n2(j)i  =  -  — 1— ^  c.g.s.  units.  (150) 

ot 

Similarly,  the  flux  produced  by  unit  current  in  B  is 

02    =    ^ 

and  the  inductance  of  B  is 


(R 

again  if  all  the  flux  produced  by  B  passes  through  A,  the  mutual 
inductance  of  B  with  respect  to  A  is 


_ 

ot 
and  is  equal  to  the  mutual  inductance  of  A  with  respect  to  B, 

3KBA  =  mAB  =  2HX. 

The  leakage  flux,  that  is,  the  flux  which  passes  out  between  the 
two  circuits  and  links  with  only  one  of  them,  has  been  assumed  to 
be  very  small.  In  this  case 


If  the  leakage  flux  is  not  negligible,  the  flux  $,  produced  by 
unit  current  in  A  can  be  divided  into  two  parts  0Ml  and  <j>8l)  Fig. 
77,  of  which  <j>Ml  interlinks  with  B  and  0Sl  forms  a  local  leakage 
circuit  about  A  and  does  not  interlink  with  B;  <f>Sl  is  the  self- 
inductive  flux  produced  by  unit  current  in  A. 

The  self-inductance  of  a  circuit  is  the  number  of  interlinkages 
with  the  circuit  of  the  flux,  produced  by  unit  current  in  it,  which 
does  not  interlink  with  any  other  circuit;  it  is  the  number  of 
interlinkages  of  the  circuit  with  the  stray  flux  or  the  leakage  flux 
produced  by  unit  current  in  itself. 

<£i  =  inductance  of  A, 

3TC  =  mutual  inductance  of  A  with  respect  to  B, 

£s=  self-inductance  of  A. 


ELECTRIC  CIRCUITS 


97 


But 

»i 

and  similarly, 


=  0Mt  +  <t>&v  and  therefore 


Turns 


O 

Ocoil 

O    B 

^v 

QTurns 


FIG.  77. — Self  and  mutual  inductance. 

When  any  magnetic  materials  are  present  the  reluctance  of  the 
magnetic  circuit  is  not  constant  but  varies  with  the  flux  density, 
that  is,  with  the  current  and  therefore  the  inductances,  the  mutual 
inductance  and  the  self-inductances  are  all  variable  quantities. 

For  any  value  of  current  i\  in  A  producing  a  flux  (f>M  linking 
B  the  rate  of  change  of  flux  generates  in  B  an  e.m.f.  which  may 
be  expressed  as 

e      =n    d$M  =  ^     dii  ,152) 

dt  dt 

and  the  mutual  inductance  of  A  with  respect  to  B  is 

3H«  =  n^-  (153) 


-p-  can,  however,  only  be  evaluated  on  the  assumption  that  the 

flux  is  proportional  to  the  current  or  that  the  reluctance  of  the 
magnetic  circuit  is  constant.  In  practical  applications  allowance 
must  be  made  for  the  errors  resulting  from  this  assumption. 

99.  Self -inductance  of  Continuous -current  Circuits.— The 
self-inductance  of  continuous-current  circuits  is  apparent  only 
when  the  current  is  increasing  or  decreasing.  The  two  most  im- 


98  ELECTRICAL  ENGINEERING 

portant  cases  are  when  the  current  is  starting  and  when  it  is 
stopping. 

Starting  of  Current.  —  When  a  constant  e.m.f.  E  is  impressed 
on  a  circuit  of  resistance  R  and  inductance  L  the  current  does  not 
immediately  reach  a  steady  value  on  account  of  the  opposing 
e.m.f.  due  to  inductance.  If  at  time  t  after  E  is  impressed  the 

current  is  changing  at  the  rate  -T  ,;  the  e.m.f.  of  inductance  is 

Tdi 
€b=     ~Ldt 

By  Lenz's  law  it  opposes  the  impressed  e.m.f.  and  is  therefore 
negative. 

The  e.m.f.  acting  on  the  circuit  is 


and  the  current  is 


therefore 

.  _  E  _  L  di 

~R~  R  dt 
or,  transposing, 

di  _  Rdt. 

E~  L' 


the  integral  of  this  is 

Rt 


i        /.       M  Kt   .   .       n 

loge  (i  -  ft)   =  -  -£  +  loge  C, 


where  loge  C  is  the  constant  of  integration.     This  reduces  to 

Rt 

" 


•    #_i_^~T> 

or  ^  =  p  +  Ge 

R 

when  t  =  0,      i  =  0,       and  C  =  —  ^ 

Substituting  this  value  the  current  is 

\t\ 

(154) 


ELECTRIC  CIRCUITS  99 

W 

When  t  =  a  and  i  =  ^>  and  the  current  has  reached  its  steady 
rt 

value,  which  .is  independent  of  the  inductance  of  the  circuit; 
calling  this  value  I  and  substituting 

(155) 


The  expression  „  is  called  the  time  constant  of  the  circuit. 

Stopping  of  Current.  —  If  the  impressed  '  e.m.f.  E  is  removed 
from  a  circuit  of  resistance  R  and  inductance  L,  when  it  is  carry- 

E 
ing  a  current  7  =  j?  and  the  circuit  is  closed  through  a  resistance 

Ri,  the  current  will  not  at  once  fall  to  zero. 

The  e.m.f.  of  inductance  at  time  t  after  the  circuit  is  closed  is 
•     .   .  ,.  di 

et=  -Ldt 

and  it  opposes1  the  decrease  of  current. 

The  currejit  in  the  circuit  at  this  instant  is 

T  di 
Ldt 

1T+TS' 

or,  transposing, 

A  (R  +  fli)  . 

i   "  L       *** 

the  integral  of  this  is 

logei=  - 


T  *        I        *v'6     v-/  ) 

where  log  C  is  the  constant  of  integration.     This  reduces  to     , 

i  =  Ce~ L      '» 
when  t  =  0,  i  =  7,  and  C  =  7. 

Substituting  this  value,  the  current  is 

i=Ie  fj  (156) 

or 

<;-i'~^'-  <167) 

The  e.m.f.  generated  in  the,  coil,  at  the  time  t,  is 

"   /  T>       I        7")    \     7vT  -^  k        I       •*•  **  1       "™"  , t  /  -t  P*  O  \ 

66  =  i  (ti  +  /ti)  =  E, ^ e         L  (158) 


100  ELECTRICAL  ENGINEERING 

When  t  =  0,  the  generated  e.m.f.  is 

eb  =  #7^^-;  (159) 

therefore  the  e.m.f.  generated  in  the  circuit  is  greater  than  the 

DID 

impressed  e.m.f.  in  the  ratio  — ^ — -• 

If,  when  the  impressed  e.m.f.  is  removed,  the  circuit  is  short- 
circuited,  RI  =  0  and  the  generated  e.m.f.  is 

eb  =  E, 

that  is,  the  e.m.f.  does  not  rise. 

But  if  the  circuit  is  broken  Ri  =  &  and  eb  =  <* ,  and  danger- 
ous e.m.f s.  are  generated  in  the  circuit.  A  large  value  of  RI 
causes  the  current  to  decrease  rapidly,  but  it  also  causes  a  high 
e.m.f.  to  be  generated. 

The  energy  supplied  to  the  circuit  while  the  current  is  starting 
is 

W  =  f    Ei  dt} 

Jo 
but 

therefore, 

/»<X       /  J  '\  /»tt  f*I 

W  =    I     ( Ri  -f  L  TJ  idt  =  I     Ri2  dt  +  I    Li  di\ 
Jo    \  0*1  Jo  Jo 

f  " 

of  this   I     Ri2  dt  is  the  energy  transformed  into  heat  in  the  electric 
Jo 

C1 
circuit,  and   I   Li  di  is  the  energy  stored  in  the  magnetic  field. 

Jo 

The  energy  in  the  magnetic  field  is 

C1  I2 

WM   =  Lidi=L^ 

Jo  * 

and  is  independent  of  the  resistance  of  the  circuit. 

This  magnetic  energy  is  returned  to  the  electric  circuit  while 
the  current  is  stopping  and  prevents  it  from  immediately  falling 
to  zero.  It  is  transformed  into  heat  in  the  circuit  and  is 

W  =    ("  i2(R  +  Rl}dt, 

Jo 
but 

R  +Ri  . 


ELECTRIC  CIRCUITS  101 

therefore, 


/» 

W  =  I2(R  +  Ri)l 
Jo 
the  integral  of  this  is 


e~ 


and  W  =  L  ~  as  before. 

40 

If  the  circuit  is  broken  this  amount  of  energy  must  be  dis- 
charged and  produces  a  spark. 

100.  Example. — The  shunt  field  winding  of  a  6-pole,  125-volt,  direct-cur- 
rent generator  has  500  turns  per  pole  and  carries  12.5  amp.;  the  resistance  of 

125 

the  field  winding  is  7.5  ohms  and  the  resistance  of  the  field  rheostat  is  -r^Tc  — 

iz.o 

7.5  =  2.5  ohms;  and  the  flux  per  pole  is  6.3  X  106  lines;  determine: 

(a)  the  inductance  of  the  field  winding; 

(6)  the  time  taken  by  the  current  to  reach  one-tenth  full  value  and  one- 
half  full  value,  after  the  field  switch  is  closed ; 

(c)  the  initial  rise  of  voltage,  if  the  field  circuit  carrying  a  current  of  12.5 
amp.  is  closed  through  a  resistance  of  25  ohms; 

(d)  the  energy  stored  in  the  magnetic  field  when  the  current  is  12. J F-  amp. 
(a)  The  interlinkages  of  flux  and  turns  with  a  current  of  12.5  amp.  or 

1.25  c.g.s.  units  is 

6  X  500  X  6.3  X  106; 
therefore  the  inductance  is 

6  X  500  X  6.3  X  106 
= 1~2^ —       — -  =  15.1  X  109  c.g.s.  units, 

or 

L  =  15.1  henrys. 

(6)  At  time  t  after  the  field  switch  is  closed  the  current  is,  by  equation 

(155), 


where  /  is  the  final  value  of  the  current  =  12.5  amp.  and  R  is  the  resistance 
of  the  field  circuit  including  the  rheostat  =  10  ohms;  therefore, 


/  _!OL\ 

i  =  12.5  ll  -  e     15-V  . 


Let  time  for  the  current  to  reach  one-tenth  of  full  value  or  1.25  amp.  be 
ti  sec.,  then 

1.25  =  12.5  (1  -  e-0-66'!) 
and 

ti  =  0.16  sec. 

Let  time  for  the  current  to  reach  one-half  value  or  6.25  amp.  be  <2  sec.,  then 

6.25  =  12.5  (1  -  e-°-66'2) 
and 

tz  =  1.05  sec. 


102  ELECTRICAL  ENGINEERING 

(c)  If  the  field  circuit  carrying  a  current  of  12.5  amp.  is  closed  through  a 
resistance  of  25  ohms,  the  initial  value  of  the  back  voltage  due  to  inductance 
is,  by  equation  (159), 

eb  =  125  10  +  25  =  437.5  volts. 

(d)  When  the  current  is  12.5  amp.  the  energy  stored  in  the  magnetic  field 
is 


W  =  L-  =  15.1  X  ^^  =1,460  watt-sec. 

101.  Inductance  of  Parallel  Conductors. — Determine  the  in- 
ductance of  the  circuit  in  Fig.  78  consisting  of  two  parallel  wires 
A  and  B  carrying  equal  currents  but  in  opposite  directions. 

R  =  radius  of  each  wire  in  centimeters. 

D  =  distance  between  centers  in  centimeters. 

A  B 


D-3- 

• — D 


FIG.  78. — Inductance  of  parallel  wires. 

When  a  current  7  c.g.s.  units  flows  in  the  wires  the  field  intensity 
at  the  point  P  distant  x  cm.  from  A  and  D  —  x  cm.  from  B  is 

27  •         27 

3C  = —  +  yr—  —dynes,  by  equation  (104), 
x       LJ  ~~  x  •  . 

and  the  flux  density  is 

27  27 

„          (B  =  5C  =  - — h  y; —    —  lines,  per  square  centimeter; 

X          LJ         X 

the  flux  in  a  section  of  width  dx  and  length  1  cm.  is 
d<f>  =  (8  dx  lines. 

The  flux  between  the  wires  per  centimeter  length  produced  by 
the  current  in  one  wire  is 

D 


-  J« 


x  '..~.~*  ~  R 


27  log 

R 

and  the  flux  of  self -inductance  of  the  circuit  per  centimeter  length 

is 

JT1      D 

^  =  2<£i  =  47  log  ~> 


ELECTRIC  CIRCUITS 


103 


Xbut  only  half  of  this  Q.UX  surrounds  each  wire  and  therefore  the 
inductance  of  each  wire  per  centimeter  length  is 

4> 


(160) 


2  j[) 

I/i  =  -j  —  2  log  -j5  c.g.s.  units. 

The  flux  inside  the  conductor  has  been 
neglected,  but  its  inductive  effect  can  be 
calculated  very  easily  if  it  is  assumed  that 
the  current  is  distributed  uniformly  over 
the  section  of  the  wire.  In  Fig.  79  the 
section  of  wire  A  is  shown  enlarged. 

The  flux  density  at  distance  x  cm.  from 
the  center  of  A  is 

(g  =  ?L'  FIG.  79.— Flux  inside  a 

X  '  conductor. 

where  I'  is  the  current  inside  the  radius  x  cm.  and  its  value  is 
/'  =  /  -~2  c.g.s.  units. 

The  flux  in  the  ring  of  radius  x,  width  dx  and  length  1  cm.  is 

.  ,       21'  2Ix  ,    .. 

d<f>    =  —  dx  =  -pg-  dx  lines; 

this  flux  surrounds  only  the  current  F  and  is  equivalent  to  a 
smaller  flux  surrounding  the  current  I,  of  value 

,  x2        2lx* 

and  the  flux  equivalent  to  the  flux  inside  the  whole  section  is 

f"  f'2I     .. 

02  ==         d<p  ==          -J-)A.  x  dx 
Jo  Jo    M 

21  rx*Y      Ir 
=  ~R*    T        =  2  "nes> 

and  the  inductance  per  centimeter  due  to  the  flux  inside  the  con- 
ductor is  02       i 

£2  =  -7-  =  o  c-g-s-  units.  (161) 

/        z 


The  inductance  of  each  wire  per  centimeter  is  therefore 


£  =  £1  +  £2  =  2  log^  +  ^c.g.s.  units, 


(162)      V 


and  the  inductance  per  mile  of  wire  is 

L  =  2.54  X12X  5,280  (2  X  2.303  loglo  ^  + 

=  ^0.74  Iog10  ^  +  0.0805)  10~3  henrys. 


10'9  henrys 

(163) 


CHAPTER  IV 
ELECTRIC  CIRCUITS  (Continued) 

ALTERNATING-CURRENT  CIRCUITS 

102.  The  Sine  Wave  of  Electromotive  Force  and  Current. — If 

the  coil  in  Fig.  80  rotates  with  a  constant  angular  velocity  in 
a  uniform  magnetic  field  between  the  poles  N  and  S,  an  alter- 
nating e.m.f.  will  be  generated  between  the  terminals  t  and  t'. 
Referring  to  the  figure  it  can  be  seen  that  in  position  (1)  the  e.m.f. 
generated  in  the  coil  is  zero,  since  the  conductors  forming  it  are 
moving  parallel  to  the  lines  of  magnetic  flux  and  therefore  are 
not  cutting  them;  in  position  (2)  it  is  positive  and  increasing;  in 


Rotation 


FIG.  80. — Generation  of  an  alternating  e.m.f. 

(3)  it  is  positive  and  has  reached  its  maximum  value  since  the 
conductors  are  cutting  perpendicularly  across  the  flux;  in  (4)  it  is 
positive  and  decreasing;  in  (5)  it  is  zero;  in  (6)  it  is  negative  and 
increasing;  in  (7)  negative  and  maximum;  in  (8)  negative  and  de- 
creasing and  in  (1)  is  zero  again,  having  gone  through  one  com- 
plete cycle.  This  cycle  is  represented  in  Fig.  80. 

If  <f>  =  the  maximum  flux  inclosed  by  the  coil,  n  =  the  number 
of  turns  on  the  coil  and  co  =  the  angular  velocity  in  radians  per 
second,  then  at  time  t  sec.  after  the  position  of  maximum  in- 
closure  the  coil  has  moved  through  angle  6  =  ut  radians,  and  the 
flux  inclosed  is 

0  =  <j>  cos  cot; 
104 


ELECTRIC  CIRCUITS  105 

the  e.m.f.  generated  in  the  coil  at  this  instant  is 

e  =  —  n  j-.  (<£  cos  co£) 
at 

=  con$  sin  ut  absolute  units 

=  ow<l>  10~8  sin  ut  volts.  (164) 

When  0  =  ~,  the  e.m.f.  has  its  maximum  value 

2 


Em  =  con$  10-8  volts, 

and  therefore  ^      .       ,  /i«c\ 

e  =  Em  sin  ut',  (165) 

and  the  e.m.f.  generated  in  the  coil  varies  as  a  sine  wave. 

The  number  of  cycles  through  which  the  e.m.f.  passes  in  one 
second  is  called  its  frequency  and  since  one  cycle  represents  360 
/electrical  degrees,  the  frequency  may  be  expressed  as 

/  =  -^  cycles,  (166) 

and  therefore  0   . 

CO    =    2-7T/. 

Substituting  this  value  of  co  in  equation  (164)  gives 
e  =  2irfn3>  10~8  sin  2irft 

=  Em  sin  6, 

where  0  =  Zirft  is  the  angle  turned  through  in  time  t  sec.  after  the 
position  of  zero  e.m.f.;  the  maximum  value  of  the  e.m.f.  is 

Em  =  2irfn3>  10~8  volts.  (167) 

In  a  two-pole  alternating-current  generator  one  cycle  corre- 
sponds to  one  revolution  of  the  coil  and 
/  =  rev.  per  sec., 

and  one  electrical  space  degree  is  equal  to  one  mechanical  space 
degree.  With  a  p-pole  alternator  the  e.m.f.  passes  through  one 
cycle  for  each  pair  of  poles  and  the  frequency  is 

/  «  ?  rev.  per  sec., 
A 

and  one  electrical  space  degree  is  less  than  one  mechanical  space 
2 

degree  in  the  ratio  -• 

If  a  non-inductive  resistance  of  R  ohms  is  connected  across  the 
terminals  of  the  coil  in  Fig.  81  an  alternating  current  will  flow  in 
the  coil  of  instantaneous  value 

Em  sin  6 


Im  sin 


106 


ELECTRICAL  ENGINEERING 


E 

where  Im  =  -^  is  the  maximum  value  of  the  current.     The  cur- 
K 

rent  and  voltage  waves  pass  through  their  zero  and  maximum 
values  together  and  are  therefore  in  phase.  They  are  represented 
by  the  two  curves  in  Fig.  81. 


FIG.  81. — Resistance  in  alternating-current  circuits. 

103.  The  Average  Value  of  a  Sine  Wave. — The  average  value 
of  the  ordinate  of  a  sine  wave  i  =  Im  sin  6  can  be  found  by  in- 
tegrating over  one-half  wave;  it  is 


1C*       . 

Jo 


sin  0  d6 


=  -f[  -  cos  $H 
=  —  [1  +  1]  =  -  Im  =  0.6377, 

TT  7T 


(169) 


that  is,  the  average  value  of  the  ordinate  of  a  sine  curve  is 

2 

-  times  the  maximum  ordinate. 

7T 

104.  The  Effective  Value  of  a  Sine  Wave. — The  effective  value 
of  an  alternating  current  is  the  value  of  continuous  current  which 
would  have  the  same  heating  effect. 

When  an  alternating  current  i  =  Im  sin  6  flows  through  a 
resistance  R}  energy  is  transformed  into  heat  at  the  instanta- 
neous rate  i2R  watts ;  the  average  rate  of  transformation  of  energy 
is 


1  C*  1  C* 

-  I    i2#  d0  =  -  I  J 

VO  ^JO 


MR 


sin2  B  d6 


-  cos  26)  d6 


ELECTRIC  CIRCUITS  107 


27T 
r     2P 

H 


27T 

'*R=PR:          (170) 


thus  the  alternating  current  i  =  Im  sin  0  has  the  same  average 
heating  effect  or  consumes  the  same  average  power  as  a  con- 

tinuous current  —  j=  ;  the  value  I  =  -~=  is  therefore  called  the 
V2  V2 

effective  value  of  the  alternating  current  i  =  Im  sin  6 

The  effective  value  of  the  alternating  e.m.f.  e  =  Em  sin  8  is 

» 

(171) 


The  effective  value  of  any  alternating  quantity  is  the  square 
root  of  the  mean  of  the  squared  instantaneous  values  taken  over 
one  complete  cycle;  it  is  equal  to  the  maximum  value  divided 
by  \/2  only  in  the  case  of  sine  waves. 

Alternating-current  voltmeters  and  ammeters  indicate  the 
effective  values  of  voltage  and  current  regardless  of  the  wave 
form. 

105.  Inductance  in  Alternating-current  Circuits.  —  When  a  cur- 
rent flows  in  a  conductor  a  magnetic  field  is  produced  in  the  space 
surrounding  it  ;  as  long  as  the  current  remains  constant  this  field 
does  not  react  on  the  electric  circuit,  but  when  the  current  varies 
the  flux  linking  with  the  circuit  also  varies  and  induces  in  the 
conductor  an  e.m.f.  opposing  the  change  in  the  current  and  con- 
sequently the  change  in  the  flux.  This  action  is  due  to  the  inertia 
of  the  niagnetic  field  and  is  analogous  to  the  action  of  the  fly- 
wheel in  mechanics.  The  inertia  of  the  flywheel  opposes  any 
change  in  speed  just  as  the  inertia  of  the  magnetic  field  opposes 
any  change  in  current.  Energy  is  stored  in  the  flywheel  as  the 
ppeed  increases  and  is  given  back  as  it  decreases  and  the  only  loss 
of  energy  is  that  due  to  friction.  Similarly  energy  is  stored  in 
the  magnetic  field  as  the  current  increases  and  is  returned  to  the 
electric  circuit  as  the  current  decreases,  and  the  only  loss  of  energy 
is  that  due  to  hysteresis  and  eddy  currents  in  the  iron  parts  of 
the  magnetic  circuit. 


108 


ELECTRICAL  ENGINEERING 


The  energy  stored  in  the  flywheel  is 

w  -  4 


(172) 


where  /  is  its  moment  of  inertia  and  co  is  its  angular  velocity. 
The  energy  stored  in  the  magnetic  field  is 

W  =—=-  watt-sec., 

where  L  is  the  inductance  of  the  circuit  in  henrys  and  i  is  the 
current  in  amperes. 

The  inductance  of  the  coil  opposes  the  change  in  current  by 
generating  a  back  e.m.f. 

di 


T 
=  "  L 


dt 


If  an  alternating  current  i  =  Im  sin  6  =  Im  sin  2-n-ft  is  flowing 
through  a  circuit  of  inductance  L  henrys  and  negligible  resist- 
ance an  e.m.f.  of  inductance  will  be  set  up 

di 


=   —  L  -r  (Im  sin  2wft) 

-  2irfLIm  cos  2irft. 
2irfLIm  sin  (2irft.  -  90) ; 


(173) 


FIG.  82. — Inductance  in  alternating-current  circuits. 

it  is  a  sine  wave  of  e.m.f.  lagging  behind  the  current  wave  by 
90  degrees  (Fig.  82). 

In  order  to  drive  the  current  through  the  circuit  an  e.m.f. 
must  be  applied  to  the  terminals,  which  at  every  instant  will  be 
equal  and  opposite  to  the  back  e.m.f.  €b',  its  instantaneous  value 

e  =  -  eb  =  2wfLIm  sin  (2irft  +  90),  (174) 

a  sine  wave  of  e.m.f.  leading  the  current  wave  by  90  degrees. 


ELECTRIC  CIRCUITS  109 

This  e.m.f.  is  consumed  by  the  inductance  of  the  circuit.     Its 
maximum  value  is 

Em  =  2irfLIm,  (175) 


where  the  term  2irfL  is  called  the  inductive  reactance  of  the  cir- 
cuit and  is  denoted  by  X.  It  is  of  the  nature  of  resistance  and 
is  expressed  in  ohms. 

Thus,  in  a  circuit  of  reactance  X  and  negligible  resistance  the 
impressed  e.m.f.  is  90  degrees  ahead  of  the  current,  or  the  cur- 
rent lags  90  degrees  behind  the  impressed  e.m.f. 

106.  Resistance  and  Reactance  in  Series.  —  If  a  circuit  con- 
tains a  resistance  R  and  a  reactance  X  in  series  and  carries  an 
alternating  current  i  =  Im  sin  2irflj  determine  the  value  of  the 
impressed  e.m.f.  and  its  phase  relation  with  the  current  (Fig.  83). 


Curve    ©,  6 

Curve  (4), 
Curve 


Curve 
FIG.  83. — Resistance  and  reactance  in  series. 


To  drive  a  current  i  =  Im  sin  2irft  (curve  1)  through  a  resist- 
ance R  an  e.m.f.  is  required  equal  to 

eR  =  iR  =  ImR  sin  2irft  (curve  2), 

a  sine  wave  in  phase  with  the  current  with  a  maximum  value  ImR' 
The  inductance  of  the  circuit  sets  up  a  back  e.m.f., 

eb  =  -  Ljt  =  27r/L7wsin  (2ir/*  -  90)  (curve  3), 

a  sine  wave  lagging  90  degrees  behind  the  current  with  a  maxi- 
mum value  2irfLIm  =  ImX. 


110  ELECTRICAL  ENGINEERING 

To  overcome  this  back  e.m.f.  due  to  inductance  an  e.m.f.  must 
be  impressed  on  the  circuit  equal  and  opposite  to  65. 

T  di 
ex  =  -eb  =  Ldl 

=  2irfLIm  sin  (27r#  +  90) 

=  ImX  sin  (2ttft  +  90)  (curve  4). 

This  is  a  sine  wave  90  degrees  ahead  of  the  current  wave,  with  a 
maximum  value  ImX  and  is  the  e.m.f.  consumed  by  the  reactance 
in  the  circuit. 

The  instantaneous  value  of  the  impressed  e.m.f.  (curve  5)  is 
the  sum  of  the  instantaneous  values  of  eR  and  ex,  thus 

e   =  eR  +  ex  =  Ri  +  L^  (176) 

=  Im  R  sin  2irft  +  ImX  sin  (2nft  +  90) 
=  Im  R  sin  2wfl  +  ImX  cos  2irft 
=  Im  (R  sin  B  +  X  cos  6) 


=  Im  Z  sin  (0  +  0),  (177) 

where  Z  =  \/  R2  -f-  X2  is  called  the  impedance  of  the  circuit  and 
is  expressed  in  ohms  and  $  is  the  angle  of  lead  of  the  impressed 
e.m.f.  relative  to  the  current.- 

X  R 

sm  <t>  =  ^===  and  cos  0  = 


The  impressed  e.m.f.  is  therefore  a  sine  wave  leading  the  cur- 
rent by  an  angle  <£.     Its  maximum  value  is 


Em  =  ImV  R2  +  X2  =  ImZ,  (178) 

and  its  effective  value  is 

J.  _  I^Z 
?       V2       V2      IZ' 

where  I  is  the  effective  value  of  the  current. 
The  component  of  E  in  phase  with  the  current  is 

T? 

El  =  IR  =  -  R  =  E  cos  0; 
& 

and  the  component  of  E  90  degrees  ahead  or  in  quadrature  ahead 
of  the  current  is 

E2  =  IX  =      X  =  #sin<. 


ELECTRIC  CIRCUITS  111 

107.  Capacity  in  Alternating-current  Circuits. — The  charge 
on  a  condenser  or  the  quantity  of  electricity  stored  in  it  is  pro- 
portional to  the  difference  of  potential  between  its  terminals; 
thus, 

q  =  Ce, 
where   q  is  the  charge, 

e  is  the  difference  of  potential  between  the  terminals, 
and       C  is  the  capacity  of  the  condenser. 

A  condenser  has  a  capacity  of  one  farad  when  one  coulomb 
of  electricity  stored  in  it  produces  a  difference  of  potential  of 
one  volt  between  its  terminals. 

If  an  alternating  e.m.f.  e  =  Em  sin  2irft  is  impressed  on  the 
terminals  of  a  condenser  of  capacity  C  farads,  a  current  i  flows. 
At  any  time  t  the  charge  on  the  condenser  is 

q  =  Ce, 

but  the  charge  is  the  amount  of  electricity  which  has  flowed  into 
the  condenser  and  its  value  is 


=  ft  A 

Jo 


and  therefore 

fi  dt  =  Ce. 

Differentiating  with  respect  to  t  gives 

and  substituting  the  value  of  e, 

i  =  Cj  (Emsm2irft) 

=  2irfCEm  cos  2irft 

=  2irfCEm  sin  (27T/Y  +  90),  (179) 

and  thus  the  current  flowing  into  the  condenser  is  a  sine  wave 
leading  the  impressed  e.m.f.  by  90  degrees.  Its  maximum  value 
is 


(180) 

-A-c 

the  condensive  reactance  of  the 
and  is  expressed  in  ohms. 


where  Xc  =  ~  ^  is  called  the  condensive  reactance  of  the  circuit 


112 


ELECTRICAL  ENGINEERING 


In  Fig.  84  curve  1  represents  the  impressed  e.m.f.  e  —  Em 
sin  2irft  and  curve  2  the  current  i  =  Im  sin  (2irft  +  90).  Be- 
tween the  points  a  and  b  the  current  is  positive  and  the  e.m.f.  is 
increasing;  from  6  to  c  the  current  is  negative  and  the  e.m.f.  is 
decreasing. 


FIG.  84. — Capacity  in  alternating-current  circuits. 

108.  Resistance  and  Condensive  Reactance  in  Series. — If  an 

alternating  current  i  =  Im  sin  2wft  flows  in  a  circuit  consisting 
of  a  resistance  R  in  series  with  a  condenser  of  capacity  C  farads 

and  reactance  of  Xc  =  ,   ,  ~,  ohms,  determine  the  magnitude  of 

the  impressed  e.m.f.  and  its  phase  relation  with  the  current 
(Fig.  85). 


Curve  ©    <?=tR  +  7f 


4 WW/,/ 

R 


FIG.  85. — Resistance  and  condensive  reactance  in  series. 

The  impressed  e.m.f.  consists  of  two  components;  one  is  re- 
quired to  drive  the  current  i  (curve  1)  through  the  resistance  of 
the  circuit;  it  is  eR  =  iR  (curve  2),  in  phase  with  i\  the  other 
component  is  the  e.m.f.  required  at  the  terminals  of  the  con- 
denser or  the  e.m.f.  consumed  by  the  condensive  reactance  of 

the  circuit;  it  is  exc  =  ^  (curve  3)  and  was  shown  to  lag  90  degrees 
behind  the  current,  Art.  107. 


ELECTRIC  CIRCUITS  113 

The  total  impressed  e.m.f.  is  (curve  4) 
e  =  eK  +  exc  (181) 


c 

1        f  T 

sin  2irft  +  — -  I  Im  sin 
CJ 
I 
™    cos 

=  Im  (R  sin  2irft-Xc  cos  2ir/0 


/ 
=  .R/w  sin  27rft  -  ^^  cos  2irft 


sn 


sin  (2irft  -  <£),  (182) 

where  Z  =  \/R2  +  ^c2  is  the  impedance  of  the  circuit  in  ohms, 
and  0  is  the  angle  of  lag  of  the  impressed  e.m.f.  behind  the  cur- 

rent;  sin*  =  'y|==,  cos  *  = 


109.  Resistance,  Inductance  and  Capacity  in  Series.  —  If   a 

current  i  =  Im  sin  27T/Y  flows  in  a  circuit  consisting  of  a  resistance 
E,  inductance  L  ar;d  capacity  C  connected  in  series,  the  impressed 
e.m.f.  is 

e  =  eR  +  ex  +  exc,  (183) 

Where  eR  =  iR  is  the  e.m.f.  consumed  by  the  resistance  of 

the  circuit, 

&x  =  L  -r  is  the  e.m.f.  consumed  by  the  inductance  of 
ctt 

the  circuit, 

exc.  =  -79  is  the  e.m.f.  consumed  by  the  capacity  of 
the  circuit. 

Assuming  that  the  impressed  e.m.f.  leads  the  current  by  angle 
0,  its  equation  is 

e  =  Em  sin  (27T/Z  +  0),  (184) 

and 


Em  sin  (2irft  +  0)  =  Ri  +  L  ~  +  ± 


(185) 


114  ELECTRICAL  ENGINEERING 

Differentiating  with  respect  to  t, 

' 


2*fEm  cos  (toft  +  0)  =  R    '  +  L      '  +  (186) 


but 

cfc  r72?" 

i  =  /„  sin  2^,  ^  =  2,r/Zm  cos  ar/<,  ^  =  -  (21r/)27m  sin 

when  (  -  0,  t  =  0,  |  =  2*//m,  g    =  0, 

when 


=    ™  cos  0,      =    7rw  sn  ^  ~=  -  tof)*Im  cos  0. 

Substituting  these  values  in  equation  (186),  when  t  =  0, 
2irfEm  cos  0  =  2irfRIm. 


and 


90-0 
when  i  =  —          ? 

ZTT; 

0  =  27r/#/m  sin  0  -  (2irf)2LIm  cos  0  +  ^  cos  0, 
and 

72  sin  0  =  27T/L  cos  0  —  cos  0, 

therefore, 

'       ••     t    „ 

tan  0  = 


=  ^A  .."  (188) 

sin  0  =  X  -  Xc        . 

VR*  +  -(x  -  xcy 

7? 

cos  0  =  —T-  >  (190) 

/—  f  (X  -  Xc)2 


and  from  equation  (162) 

T7T  13 

•"  Wl  ^^ 


where 

z  = 


is  the  impedance  of  the  circuit. 


ELECTRIC  CIRCUITS 


115 


The  impressed  e.m.f.  is  therefore  a  sine  wave  of  maximum 

value  Em  =  ImZ  =  Im  VR*  +  (X  -  Xc)2,  and  leads  the  current 

X  —  X 
wave  by  an  angle  <£  =  tan"1 ^ — ~* 

If  Xc  >  X,  <£  is  negative  and  the  impressed  e.m.f.  lags  behind 
the  current. 

If  Xc  =  X,  0  =  0  and  the  e.m.f.  is  in  phase  with  the  current, 
and  the  impedance  of  the  circuit  is  Z  =  R. 

110.  Vector  Representation  of  Harmonic  Quantities. — A  sine 
wave  may  be  obtained  from  a  circular  locus  as  shown  in  Fig.  86. 


s 

)=wt 


\/ 


\_« 


s£z 


FIG.  86. — Sine  wave. 

The  radius  of  the  circle  is  the  maximum  value  of  the  sine  func- 
tion and  is  called  its  amplitude.  If  the  radius  vector  is  rotated 
at  uniform  angular  velocity  o>  in  the  counter-clockwise  direction, 
its  vertical  projection  at  any  time  t  is 

y  =  R  sin  co£  =  R  sin  6, 

where  time  is  measured  from  the  horizontal.     Plotting  the  values 
of  y  on  a  base  of  angle  6  for  a  complete  revolution  of  the  radius 
vector  gives  the  cycle  shown. 
To  represent  the  sine  function 

e  =  Em  sin  0, 

a  circle  of  radius  Em  is  taken  and  the  vertical  projections  of  the 
revolving  vector  are  plotted  on  base  of  angle  0  (Fig.  87 (a)). 
To  represent  a  sine  function 

i  =  Im  sin  (B  —  4>), 

a  circle  of  radius  Im  is  taken,  but  since  i  does  not  pass  through 
zero  until  angle  6  =  <f>,  the  sine  wave  is  displaced  to  the  right  by 
angle  0  as  shown  and  the  e.m.f.  wave  ^=  Em  sin  0  leads  the  cur- 
rent wave  i  =  Im  sin  (6  —  0)  by  angle  <f>.  Instead  of  using  com- 
plete circles  to  represent  sine  waves,  their  radii  Em  and  Im  can 


116 


ELECTRICAL  ENGINEERING 


be  used  as  in  Fig.  87(6)  or  since  alternating  quantities  are  repre- 
sented by  their  effective  values  instead  of  their  maximum  values 
the  two  vectors  OE  and  01,  Fig.  87  (c),  are  used  to  represent  the 

W 

two  sine  waves  in  Fig.  87 (a).     OE  =  E  =  — j=  is  the  effective 

V2 

value  of  the  e.m.f .  e  =  Em  sin  6,  and  01  =  I  =  — ^  is  the  effect- 

V2 

ive  value  of  the  current  i  =  Im  sin  (0  —  <£).  The  vector  01  is 
behind  the  vector  OE  by  angle  <£  and  so  indicates  the  relative 
phase  of  the  two  quantities.  The  counter-clockwise  direction  is 


FIG.  87. — Vector  representation  of  harmonic  functions. 


taken  as  the  direction  of  advance  in  phase  since  it  was  adopted 
at  the  International  Electrotechnical  Congress  at  Turin  and  is 
now  standard. 

111.  Power  and  Power  Factor. — In  continuous-current  cir- 
cuits the  power  consumed  is  the  product  of  the  impressed  e.m.f. 
and  the  current,  and  is- 


P  =  El  watts. 


(192) 


In  alternating-current  circuits  the  power  varies  with  time;  its 
instantaneous  value  is 


ei  watts, 


(193) 


where  e  and  i  are  the  instantaneous  values  of  the  e.m.f.  and  cur- 
rent in  the  circuit. 

Two  cases  will  be  considered,  first,  when  the  current  and  e.m.f. 
are  in  phase  and,  second,  when  the  current  lags  behind  the  e.m.f. 

Case  I. — If  an  e.m.f.  e  =  Em  sin  6  is  impressed  on  the  terminals 
of  a  circuit  of  resistance  R  a  current  i  =  Im  sin  6  will  flow  in 
phase  with  the  e.m.f. 


ELECTRIC  CIRCUITS  117 

The  instantaneous  power  in  the  circuit  is 

p  =  ei  =  Emlm  sin2  0  =  ^^  (1  -  cos  26).  (194) 

The  values  of  e,  i  and  p  for  a  complete  cycle  are  plotted  in  Fig.  88. 
The  power  varies  with  twice  the  frequency  of  the  current  from 
0  to  Emlm,  but  never  becomes  negative.  The  area  beneath  the 
power  curve  represents  the  energy  consumed  in  the  circuit  during 
one  cycle. 


if 

=  -I  P 

"  f/  U 


FIG.  88. — Power  in  a  non-inductive  circuit. 

The  average  power  is 
P 

= l  ("T^  (i  -  c°«  ^ de 

=  ErnL.nl- ^        sin  26 

EmL 


27T 
EJ[, 


M 


Em      Im 


(195) 


2  V2    V2 

therefore,  the  average  power  in  an  alternating-current  circuit  is 
equal  to  the  product  of  the  effective  values  of  the  e.m.f.  and  the 
current,  if  they  are  in  phase. 
Since 

P  =  El 
and 

E  =  IR,      .' 
therefore, 

P  =  PR,  (196) 


118 


ELECTRICAL  ENGINEERING 


and  the  power  is  equal  to  the  square  of  the  effective  value  of  the 
current  multiplied  by  the  resistance. 

Case  II. — If  an  e.m.f.  e  =  Em  sin  6  is  impressed  on  a  circuit 
containing  a  resistance  R,  and  an  inductive  reactance  X,  the  cur- 
rent will  lag  behind  the  e.m.f.  by  an  angle  4>  =  tan-1  TT  and  its 

instantaneous  value  will  be 

i  =  Im  sin  (0  —  0). 
The  instantaneous  power  in  the  circuit  is 

p  =  ei  =  Emlm  sin  B  sin  (B  —  0) 
Emlm 


cos  0  —  cos  (2  e  — 


(197) 


The  values  of  e,  i  and  p  for  one  cycle  are  plotted  in  Fig.  89. 
The  maximum  value  of  power  is  ™  m  (1  +  cos  0),  which  is  less 

than  in  Case  I,  and  the  power  curve  falls  below  the  base  twice  in 
each  cycle.  The  energy  consumed  per  cycle  is  the  difference 
between  the  positive  and  negative  areas  intercepted  by  the  power 


curve. 


FIG.  89. — Power  in  an  inductive  circuit. 

The  average  power  is 

P  =-  Fp  dd 

*  Jo 

(•* 

cos0  —  cos  (20  —  0)[d0 
sin  (20  -_0)T 

,    sin  (- 


Eml, 

2*       J0 

Emlm  r 

2x 


27T 


0 • cos  0  — 

7T  •  COS  0  — 


Emlm  r  sin(- 


cos 


2 

EI  cos  0; 


(198) 


ELECTRIC  CIRCUITS 


119 


therefore,  the  average  power  is  the  product  of  the  effective  values 
of  the  e.m.f.  and  current  multiplied  by  the  cosine  of  the  angle  of 
phase  difference  between  them. 
From  the  vector  diagram,  in  Fig.  89,  it  is  seen  that 

E  cos  <f>  =  Ei  =  IR, 
and  therefore  the  power  is 

p  =  EJ  =  PR, 

and  is  equal  to  the  active  or  in-phase  component  of  the  e.m.f. 
multiplied  by  the  current  or  is  equal  to  the  square  of  the  current 
multiplied  by  the  resistance  of  the  circuit  as  found  in  equation 


FIG.  90. — Power  in  an  inductive  reactance. 


FIG.  91. — Power  in  a  condensive  reactance. 

196;  thus,  all  the  power  consumed  in  the  circuit  is  consumed  by 
the  resistance.  Reactance  or  self-inductance  does  not  consume 
power  since  the  energy  stored  while  the  current  is  increasing  is 
given  back  while  it  is  decreasing  and  the  e.m.f.  consumed  by  self- 
inductance  is  a  wattless  e.m.f.  Similarly  condensive  reactance 
does  not  consume  power  since  the  energy  stored  while  the  e.m.f. 
is  increasing  is  given  back  to  the  circuit  while  the  e.m.f.  is  decreas- 
ing and  the  e.m.f.  consumed  by  condensive  reactance  is  wattless. 
These  results  are  illustrated  in  Fig.  90  and  Fig.  91. 


120  ELECTRICAL  ENGINEERING 

In  Fig.  90  are  plotted  the  values  of  e,  i  and  p  for  an  inductive 
circuit  without  resistance  in  which  the  current  lags  90  degrees 
behind  the  impressed  e.m.f.  The  power  curve  cuts  off  equal 
areas  above  and  below  the  base  line  and  therefore  the  average 
power  is  zero.  The  area  below  the  line  represents  the  energy 
given  back  by  the  magnetic  field  while  the  current  is  decreasing 
from  its  maximum  value  Im  to  zero  and  the  area  above  the  line 
represents  the  energy  stored  in  the  magnetic  field  while  the  cur- 
rent is  increasing  again  to  its  maximum  value.  The  amount  of 

energy  in  each  case  is  L  -~-  watt-sec. 

In  Fig.  91  are  plotted  the  values  of  e,  i  and  p  for  a  circuit  con- 
taining a  condensive  reactance  but  without  resistance.  The  cur- 
rent leads  the  impressed  e.m.f.  by  90  degrees  and  the  average 
power  is  again  zero,  so  that  no  energy  is  consumed  in  the  circuit. 
The  positive  area  cut  off  by  the  power  curve  represents  the  energy 
stored  in  the  electrostatic  field  of  the  condenser  while  the  e.m.f. 
is  increasing  and  the  negative  area  represents  the  energy  re- 
turned to  the  circuit  while  the  e.m.f.  is  decreasing.  The  maxi- 
mum amount  of  energy  in  each  case  is  C  —~-  watt-sec,  where 
Em  is  the  maximum  e.m.f. 
R 

\M 

E'       - 


E  E.1X  ^E  ^E!=IX  A^>* 

=P  Kinfh    \^^        \\ 


S 


=ESin0  \ 


I     -£ * >L^i 

Ei=IR  =  E  Cos  0 
(a)  (6)  (c) 

FIG.  92. — Power  in  alternating-current  circuits. 

Fig.  92  illustrates  various  methods  of  representing  the  power 
in  a  circuit;  in  (a)  it  is  the  product  of  the  impressed  e.m.f.,  the 
current  and  the  cosine  of  the  angle  of  phase  difference  between 
them, 

P  =  E  X  I  X  cos  0;  (199) 

in  (6)  it  is  the  product  of  the  current  and  the  in-phase  component 
of  the  e.m.f., 

P  =  /X#cos0  =  /X#i  =  PR',  (200) 

in  (c)  it  is  the  product  of  the  e.m.f.  and  the  in-phase  component 
of  the  current, 

P  =  E  X  /  cos  0  (201) 


ELECTRIC  CIRCUITS 


121 


The  apparent  power  in  a  circuit  is  the  product  of  the  impressed 
e.m.f .  and  the  current  and  is  expressed  in  volt-amperes  or  kilovolt- 
amperes. 

The  power  factor  of  a  circuit  is  the  ratio  of  the  true  power  to 
the  apparent  power  and  is 


P        El  cos  <t> 

EI  =    ~ET~=COS 


(202) 


15 


0  10  20  30  40  50  60  70  80  90 

Per  Cent  Power  Factor  -  In-Phase  Component  -  Cosine 
FIG.  93. 

therefore,  the  power  factor  is  the  cosine  of  the  angle  of  phase 
difference  between  the  current  and  the  impressed  e.m.f.;  it  is 
usually  expressed  in  per  cent,  and  may  be  either  leading  or 
lagging. 

The  sine  of  the  angle  of  phase  difference  between  the  current 
and  the  impressed  e.m.f.  is  called  the  reactive  factor  of  the 
circuit.  « 

When  the  current  is  in  phase  with  the  e.m.f.,  the  power  factor 
is  unity  or  100  per  cent,  and  the  reactive  factor  is  zero. 


122 


ELECTRICAL  ENGINEERING 


When  the  current  leads  the  e.m.f .  by  30  degrees  the  power  fac- 
tor is  cos  30  =  0.866  =  86.6  per  cent,  leading  and  the  reactive 
factor  is  sin  30  =  0.500  =  50  per  cent. 

When  the  current  lags  60  degrees  behind  the  e.m.f.  the  power 
factor  is  cos  60  =  0.500  =  50  per  cent,  lagging  and  the  reactive 
factor  is  sin  60  =  0.866  =  86.6  per  cent. 

When  the  power  factor  is  90  per  cent,  the  reactive  factor  is 

*Vl  -  O92  =  0.436  =  43.6  per  cent. 
When  the  power  factor  is  99  per  cent,  the  reactive  factor  is 

Vi  -  Q.992  =  0.141  =  14.1  per  cent. 

From  this  last  example  it  may  be  seen  that  very  great  care 
must  be  exercised  in  determining  reactive  factors  from  power- 
factor  readings  when  the  power  factor  is  near  unity;  an  error  of  1 
.per  cent,  in  reading  the  current,  e.m.f.  or  power  might  cause  a 
reactive  component  of  14.1  per  cent,  to  be  missed  entirely. 

These  and  similar  problems  may  be  solved  quickly  by  using 
the  chart  in  Fig.  93. 

112.  Examples. — 1.  If  an  alternating  e.m.f.  of  effective  value  E  is  im- 
pressed on  a  non-inductive  circuit  of  resistance  R,  a  current  7  will  flow  in 
phase  with  e.m.f.,  where 

-I 

The  vector  diagram  is  shown  in  Fig.  94. 
I  I 


•R        E 


FIG.  94. 


FIG.  95.  FIG.  96.  FIG.  97. 


FIG.  98. 


2.  If  an  alternating  e.m.f.  E  is  impressed  on  a  circuit  of  reactance  X  and 

V 

negligible  resistance,  a  current  7  =  —  will  flow  lagging  90  degrees  behind  the 

X 

e.m.f.  (see  Fig.  95). 

3.  If  an  alternating  e.m.f.  E  is  impressed  on  a  circuit  of  resistance  R  and 
reactance  X,  a  current  7  will  flow  lagging  behind  the  e.m.f.  by  angle  <£,  where 


tan*  =      - 


ELECTRIC  CIRCUITS 


123 


The  vector  diagram  for  the  circuit  is  shown  in  Fig.  96.  The  e.m.f.  con- 
sumed in  the  resistance  is  EI  =  IR  volts  in  phase  with  the  current  and  is 
represented  by  the  vector  OEi. 

The  e.m.f.  consumed  by  the  reactance  is  E2  =  IX  volts  leading  the  cur- 
rent by  90  degrees  represented  by  OEZ.  The  impressed  e.m.f.  is  OE  =  E 
and  is  the  vector  sum  of  EI  and  Ez',  therefore 


E 


=  iz, 


where 


Z  =  \/R2  +  X2  is  the  impedance  of  the  circuit. 


4.  Fig.  97  shows  the  vector  diagram  of  a  circuit  containing  a  condenser  of 
reactance  Xc,  when  an  alternating  e.m.f .  E  is  impressed  on  its  terminals  and 

ET 

a  current  I  =  -~-  flows  through  it  leading  the  e.m.f.  by  90  degrees. 

Ac 

6.  Fig.  98  shows  the  diagram  for  the  same  circuit  with  a  resistance  R 
added  in  series. 

The  e.m.f.  consumed  by  the  resistance  is  OEi  =  EI  =  IR,  in  phase  with 
01  =  /. 

The  e.m.f.  consumed  by  the  reactance  is  OE2  =  E2  =  IXC,  lagging  90 
degrees  behind  OI. 

The  impressed  e.m.f.  is 

OE  =  E  =  \/Ei2  +  E22  =  I\/R2  +  Xe2  =  IZ, 

lagging  behind  OI  by  angle  </>,  where 

.       E2      IX.      Xe 


T 


t: 


FIG.  99. 

6.  If  an  alternating  e.m.f  .  E  is  impressed  on  a  circuit  containing  a  resist- 
ance R)  an  inductive  reactance  X  =  2irfL  and  a  condensive  reactance  Xe  = 

connected  in  series,  determine  the  magnitude  and  phase  relation  of  the 


current  and  draw  the  vector  diagram  for  the  circuit  (Fig.  99). 

01  =  I  is  the  current  taken  as  horizontal. 

OEi  =  EI  =  IR  is  the  e.m.f.  consumed  by  the  resistance  and  is  in  phase 
with  7. 

OEZ  =  E2  =  IX  is  the  e.m.f.  consumed  by  the  inductive  reactance  and 
leads  01  by  90  degrees. 

OE3  =  EZ  =  IXe  is  the  e.m.f.  consumed  by  the  condensive  reactance  and 
lags  behind  01  by  90  degrees. 


124 


ELECTRICAL  ENGINEERING 


OE  =  E  =  IZ  is  the  e.m.f.  impressed  on  the  circuit,  or  the  e.m.f.  con- 
sumed by  the  impedance  Z.  It  is  the  vector  sum  of  the  three  components 
Eif  E2  and  E3  and  leads  the  current  by  angle  0;  therefore, 


and 


z  = 


+  (X  -  Xc) 


(X  -  xc 


where  L  is  the  inductance  of  the  circuit  in  henrys,  C  is  the  capacity  in  farads 
and  /is  the  frequency  of  the  impressed  e.m.f. 

The  angle  of  phase  difference  between  the  e.m.f.  and  current  is  </>,  where 


tan  0  = 


El 


IR 


X  -  Xc 

TT^ 


and 


If  X  >  Xc  the  current  lags  behind  the  e.m.f.; 

if  X  =  Xc  the  current  is  in  phase  with  the  e.m.f.; 

if  X  <  Xc  the  current  leads  the  e.m.f. 


1  R 
E      III 

JX 

^70 

J              tf 

^^  I 

FIG.  100. 

7.  An  alternating  e.m.f.  E  is  impressed  on  the  terminals  of  the  circuit  in 
Fig.  100  consisting  of  a  resistance  R  and  an  inductive  reactance  X  in  parallel. 
The  main  current  I  has  two  components, 


and 


ET 

/i  =  -~,  in  phase  with  E, 


Tjl 


/2  =  ^,  90  degrees  behind  E. 
From  the  vector  diagram 


71*  +  /22  =  #\/-^-2+^-2 


E 


VR*  +  x* 

RX 


and  lags  behind  E  by  an  angle  <f>,  where 

tan      _  h  =  E/X  -  % 

The  impedance  of  the  circuit  is 

„       E  RX 




1    VR*  +  x2 

8.  In  Fig.  101  a  third  branch  is  connected  in  parallel  with  the  two  in 
example  (7)  containing  a  condensive  reactance  Xc  >  X. 


ELECTRIC  CIRCUITS 

The  main  current  7  has  three  components, 

1 1  =  —  ,  in  phase  with  E, 

pi 
/  2  =  — ,  90  degrees  behind  E, 

1 3  =  — >  90  degrees  ahead  of  E. 

I« 

I 


125 


and 


FIG.  101. 


From  the  vector  diagram 


and  lags  behind  E  by  an  angle  <j>,  where 

72  -  73  _  1/X  -  1/X. 
7/72 


tan  <f>  = 


The  impedance  of  the  circuit  is 


FIG.   102. 

9.  Find  the  magnitudes  of  the  currents  in  the  various  parts  of  the  circuit 
in  Fig.  102  and  their  phase  relations  with  the  impressed  e.m.f. 

E 

I\  =  —y         and  lags  behind  the  impressed  e.m.f.  by  an  angle  <£i, 

v  Tti2  +  -Xi2 

Xi. 

where  tan  <^>i  =  TT  » 


7,  = 


E 


Xt 


and  leads  the  e.m.f.  by  an  angle  <£2,  where  tan 


126  ELECTRICAL  ENGINEERING 


I  =  \//i2  +  I*2  +  2/i/2  cos  (0i  +  02)     (see  vector  diagram)  and  lags 
behind  the  e.m.f.  by  an  angle  0,  where 

/i  sin  0!  -  /2  sin  02 
tan  0  =  j  —       —  ;  —  y  —  •      —  ' 
/i  cos  0i  +  /2  cos  02 

The  power  consumed  in  the  circuit  is 
>  =  El  cos  0 

=  /:2#i  +  /22#2  watts. 

113.  Numerical  Examples.  —  1.  If  an  alternating  e.m.f.  of  200  volts  at  a 
frequency  of  60  cycles  per  second  is  impressed  on  a  circuit  consisting  of  a 
resistance  of  10  ohms  in  series  with  an  inductance  of  0.1  henry  and  a  capacity 
of  100  microfarads,  (a)  determine  the  current  in  the  circuit  and  its  phase 
relation  with  the  impressed  e.m.f.,  (6)  the  e.m.f.  consumed  in  each  part  of  the 
circuit,  (c)  If  the  impressed  e.m.f.  is  maintained  constant  and  the  frequency 
is  varied  determine  the  maximum  value  of  the  current,  (d)  Plot  the  current 
and  the  various  e.m.fs.  on  a  frequency  base. 
(a)  The  inductive  reactance  of  the  circuit  is 

X  =  27T/L  =  2  X  3.14  X  60  X  0.1  =  37.6  ohms; 
the  condensive  reactance  is 

1  106 

Xc  =  =  2  X  3.14  X  60  X  100  = 


the  impedance  of  the  circuit  is 

Z  =  VR2  +  (X  -  Xc)2  =  VlO2  +  (37.6"-  2^4)2  =  15  ohms; 

and  therefore  the  current  is 

T       E       200 
J-3-15    ==13.3  amp. 

The  current  lags  behind  the  e.m.f.  by  an  angle  0,  where 

X  -Xc      37.6  -  26.4 
tan0=    -3-  ~£- 

and 

0  =  48°  18'. 

(6)  The  e.m.f.  consumed  in  the  resistance  is 

Ei  =  IR  =  13.3  X  10  =  133  volts; 
the  e.m.f.  consumed  in  the  inductive  reactance  is 

Ez  =  IX  =  13.3  X  37.6  =  500  volts; 
and  the  e.m.f.  consumed  by  the  condensive  reactance  is 

E3  =  IXC  =  13.3  X  26.4  =  350  volts. 
The  impressed  e.m.f.  is 

E  =  VES+  (Et-  'E3)2  =  A/1332  +  (500  -  350)  2  =  200  volts. 
The  vector  diagram  is  shown  in  Fig.  103. 

(c)  The  current  in  the  circuit  at  any  frequency  is 


7=1= 


If  E  is  maintained  constant  at  200  volts  and  /  is  varied,  /  varies. 
When  /  =  0,  ^  =  cc  and  /  =  0; 


ELECTRIC  CIRCUITS 


127 


when/  =  oc,  2irfL  =   oc  and  /  =  0; 


when  2irfL  = 


27T/C 


or/  = 


2  X  3.14 <y).l  X~ 
the  current  has  its  maximum  value 

/max.   =  p  =  ~7n~    =  20  amp. 

and  is  in  phase  with  the  impressed  e.m.f. 

E2«=IX— 500  Volts 


50  cycles 


L—O.l  Henry 
R=10  Ohms       X=37.6  Ohms 


E  =200  Volts 

/«=  60  Cycles  per  Sec. 


8 

'     T^ 


C  =  10'4  Farad. 
Xc"=26.4  Ohms 


E=200  Volts 
I A      0-,8°,s' 
(/ Y      >I°^-3Ampe 
/  Ei=IR=-133  Volts 

=  IXC=350  VoltB 


FIG.  103. 


The  e.m.f.  consumed  in  the  resistance  is 

Ei  =  IR  =  20  X  10  =  200  volts; 
the  e.m.f.  consumed  in  the  inductive  reactance  is 

E,  =  IX  =  20  X  2  X  3.14  X  50  X  0.1  -  628  volts; 
and  the  e.m.f.  consumed  in  the  condensive  reactance  is 

1 


Ea  =  IXC  =  20  X 


E2"IX-  628  Volts 


»  20  Amperes 


2  X  3.14  X  50  X 


E-  E!=  IR-200  Volts 


VE8-IXC  -  628  VoltB 


FIG.   104. — Resonant  circuit. 


=  628  volts. 


M.F.Cojisunied  in  the 
Inductive  Eeactance  =  e2 
Effective  Value 
E2=  028  Volts 


urrent«  i,  Effective  Value 
I»20  Amperes 


Impressed  E.M.F. - 
E.M.F.  Consumed  in 
the  Kesistnuce-  e, 
Effective  Value 
E  -  200  Volts 

E.M.F. Consumed  in 
the  Condensive 
Reactance  »  e 3 
Effective  Value  ' 
Es-628  Volts 

FIG.  105. — E.m.fs.  and  currents  in  a 
resonant  circuit. 


The  vector  diagram  for  the  circuit  is  shown  in  Fig.  104,  and  the  current  and 
e.m.f.  waves  are  shown  in  Fig.  105.     A  series  circuit  in  which  the  inductive 


128 


ELECTRICAL  ENGINEERING 


reactance  and  the  condensive  reactance  are  equal  at  a  certain  frequency  is 
said  to  be  in  a  state  of  resonance  for  that  frequency.  In  commercial  circuits 
the  capacity  is  usually  so  small  that  resonance  cannot  occur  at  ordinary 
frequencies,  but  when  any  high  frequency  e.m.fs.  are  produced  in  the  circuit 
resonance  may  occur  and  very  large  e.m.fs.  may  appear  and  break  down  the 
insulation. 

(d)  In  the  table  below  are  given  the  values  of  the  circuit  constants  for 
frequencies  from  0  to  200  cycles  per  second  and  the  corresponding  values  of 
/,  Ei,  E2,  E3  and  cos  <f>. 


Current  Lagging 


Cycles  per  Second 

FIG.  106. 


/ 

X 

Xc 

x-xe 

(X-XJ*         Z' 

,|, 

Si 

E2 

Ez 

Cos  <t> 

0 

0.0 

cc 

—   a 

oc 

oc 

oc 

0.0 

0.0 

0.0 

200.0 

0.000 

10 

6.3 

159.0 

-152.7 

23,400 

23,500 

153.0 

1.3 

13.0 

8.2 

208.0 

0.065 

40 

25.1 

39.8 

-14.7 

216 

316 

17.8 

11.2 

112.0 

281.0 

446.0 

0.560 

50 

31.4 

31.4 

0.0 

0 

100 

10.0 

20.0 

200.0 

628.0 

628.0 

1.000 

60 

37.6 

26.4 

11.2 

126 

226 

15.0 

13.3 

133.0 

502.0 

353.0 

0.670 

100 

62.8 

15.9 

56.9 

3,240 

3,340 

57.7 

3.5 

35.0 

218.0 

55.2 

0.170 

200 

125.6 

7.9 

117.7 

13,800 

13,900 

118.0 

1.7 

17.0 

213.0 

13.4 

0.085 

ec 

oc 

0.0 

oc 

oc 

oc 

K 

0.0 

0.0 

200.0 

0.0 

0.000 

These  quantities  are  plotted  on  a  frequency  base  in  Fig.  106.  J£2  reaches 
its  max.  when/  =  51.4  and  E3  reaches  its  max.  when/  =  49.2  cycles  per 
second. 

2.  If  an  alternating  e.m.f.  E  =  200  volts  at  a  frequency/  =  60  cycles  per 
second  is  impressed  on  the  circuit  in  Fig.  107,  determine  the  value  and  phase 
relation  of  the  main  current  and  the  currents  in  the  three  branches. 

The  first  branch  is  a  resistance  R  =  40  ohms;  the  second  branch  is  an 
inductance  L  =  0.1  henry  and  has  a  reactance  X  =  2irfL  =  37.6  ohms  at 
60  cycles;  the  third  branch  is  a  capacity  C  —  100  microfarads  or  10~4  farads 

and  has  a  reactance  Xc  =  0  fn  =  26.4  ohms. 


ELECTRIC  CIRCUITS 


129 


The  current  in  the  resistance  is 


E      200 
/1==        =~        = 


in  phase  with  the  impressed  e.m.f.;  the  current  in  the  inductive  reactance  is 

E        200 
/2  =  X  =  37^6  =  5'3  amp" 

90  degrees  behind  the  impressed  e.m.f.;  the  current  in  the  condensive  react- 
ance is 

E       200 


90  degrees  ahead  of  the  impressed  e.m.f. 
The  main  current  is 


/  =  V/i2  +  (/a  -  /s)2  =  V52  +  (5.3  -  7.6)2  =  5.5  amp., 
leading  the  impressed  e.m.f.  by  an  angle  <£,  where 

tan  *  =  Ap[?  =  2.3       o.46, 

J  1  O 

and  therefore 

</>  =  24°  42'. 

If  the  e.m.f.  impressed  on  the  circuit  is  maintained  constant  and  the  fre- 
quency is  varied  find  the  magnitude  of  the  main  current  when  it  is  in  phase 
with  the  e.m.f. 

The  current  at  any  frequency  is 


=   V/l2+(/2-    /3) 


Is=»7.6  Amperes 


C  =10'*  Farads 
=  0 

Xc=  26.4  Ohms 


1=  j.o  Amperes 
E=200  Volts 


1=5  Amperes 

=>  24°42' 
IjsaS.3  Amperes 


FIG.  107. 
the  angle  of  lag  of  the  current  behind  the  e.m.f.  is 


When  the  current  is  in  phase  with  e.m.f. 

0  =  0  and  X  =  Xc  or  2irfL 
the  frequency  is  therefore 


1 


2*VLC  ~  2  X  3.14  V  0.1  X  HT* 


~  5° 


130 


ELECTRICAL  ENGINEERING 


The  main  current  at  a  frequency  of  50  cycles  is 

7  =  7=?  =200       „ 
111       R       40 


the  current  in  the  inductive  reactance  is 
E  200 


X      2X3.14X50X0.1 
the  current  in  the  condensive  reactance  is 
_  E._  200 

•*3     —     -y      —    "  < 

AC  1 


-    =  6.36  amp. 


=  6.36  amp.; 


2  X  3.14  X  50  X  10-4 
and  the  current  in  the  lead  between  the  first  and  second  branches  is  zero. 

3.  If  an  alternating  e.m.f.  E  =  100  volts  is  impressed  on  a  circuit,  Fig. 
108,  made  up  of  two  parallel  branches  1  and  2,  the  first  with  resistance 
Ri  =  4  ohms  and  inductive  reactance  Xi  =  3  ohms  and  the  second  with 
resistance  Rz  =  6  ohms  and  inductive  reactance  Xz  =  8  ohms;  determine  the 
magnitude  of  the  main  current  /  and  of  the  two  branch  currents  I\  and  7  2 
and  their  phase  relations  with  the  impressed  e.m.f.  E.  The  vector  diagrain 
is  shown  in  Fig.  109. 


I        (1) 


(2) 


II 

I2< 

Ri=4 

E  =  100  Volts 

1 

R2  =  6 


1 0=10 


FIG.   108. 


1-29.9 
FIG.   109. 


The  current  in  branch  1  is 
E 

1  1 


=        100 

~  \/42  +  32 
and  it  lags  behind  the  impressed  e.m.f.  by  angle 


=  20  amp., 


The  current  in  branch  2  is 


E 


100 


/2    = 

and  it  lags  behind  the  impressed  e.m.f.  by  angle 

02  =  tan""1  ^  =  tan"1  %  =  tan"1  %    =  sin"1  %  =  cos"1  2£. 
Kz 

The  main  current  /  is  the  vector  sum  of  I\  and  /2  and  may  be  found  as 
/  =\/(Ii  sin  0i  +  /2  sin  02) 2  +  (/i  cos  0i  -f-  I z  cos 


and  it  lags  behind  the  impressed  e.m.f.  E  by  angle 


tan  • 


cos  <f>i  +  h  cos 


=,  tan-         =  tan-  0.909  =  42°  18'. 


ELECTRIC  CIRCUITS  131 

114.  Circuit  Constants. — A  continuous-current  circuit  has  two 
constants : 

impressed  e.m.f. 

resistance  R,  r  =  -  — > 

current 

~  current 

conductance  G,  g  =  .- 


impressed  e.m.f. 
and  the  conductance  is  the  reciprocal  of  the  resistance  or 


Continuous-current  circuits  also  have  inductance  and  electro- 
static capacity,  but  these  do  not  affect  the  flow  of  current  except 
at  the  instant  of  opening  or  closing  the  circuit. 

An  alternating-current  circuit  has  six  so-called  constants: 

1.  resistance  R,  r, 

2.  reactance  X,  x, 

3.  impedance  Z,  z, 

4.  admittance  Y,  y, 

5.  conductance  G,  g, 

6.  susceptance  B,  b. 

1.  The  resistance  of  a  circuit  consumes  a  component  of  e.m.f. 
in  phase  with  the  current  and  so  consumes  power.     In  circuits 
which  are  partially  inclosed  in  iron  an  alternating  magnetic  flux 
is  produced  in  the  iron  and  a  loss  of  power  occurs  due  to  hysteresis 
and  eddy  currents.     These  iron  losses  are  sometimes  included 
with  the  copper  loss  and  charged  against  the  resistance.     This 
gives  a  value  of  resistance  greater  than  the  true  ohmic  resistance 
and  is  called  the  effective  resistance  of  the  circuit.     Since  the 
hysteresis  and  eddy-current  losses  vary  both  with  the  frequency 
and  the  induction  density  in  the  iron,  the  effective  resistance  is 
not  a  constant  quantity. 

The  active  component  of  the  impressed  e.m.f.  or  the  component 
in  phase  with  the  current  is 

El  =  IR} 
and  the  resistance  is 

EI       active  component  of  impressed  e.m.f. 

K  =  -=.-  =  - 

I  current 

=  m-phase  component  of  impressed  e.m.f.  (2Q3] 

current 

2.  The  reactance  of  a  circuit  consumes  a  component  of  e.m.f. 
in  quadrature  with  the  current,  leading  in  the  case  of  circuits  of 


132  ELECTRICAL  ENGINEERING 

large  inductance  and  lagging  in  circuits  of  large  electrostatic 
capacity,  but  it  does  not  consume  any  power. 
The  inductive  reactance  of  a  circuit  is 

X  =  2-n-fL  ohms, 

where  /  is  the  frequency  of  the  impressed  e.m.f.  and  L  is  the  in- 
ductance in  henrys.  Commercial  circuits  are  operated  at  a  fixed 
frequency  and  so  /  is  constant. 

The  inductance  of  a  circuit  in  air  or  any  non-magnetic  material 
is  constant  but  in  an  iron-clad  circuit  it  varies  with  the  current, 
decreasing  as  the  current  increases  since  the  permeability  of  the 
iron  decreases  as  the  flux  density  in  it  increases. 

Since  inductive  reactance  consumes  a  component  of  e.m.f.  in 
quadrature  ahead  of  the  current  it  is  taken  as  a  positive  reactance. 

The  condensive  reactance  of  a  circuit  is 

Xc  = 


where  C  is  the  capacity  of  the  circuit  in  farads.  The  capacity  of 
a  circuit  does  not  vary  with  the  current  or  e.m.f.  and  thus 
the  condensive  reactance  is  constant  so  long  as  the  frequency  is 
constant. 

Condensive  reactance  is  taken  as  a  negative  reactance  since  it 
consumes  a  component  of  e.m.f.  in  quadrature  behind  the  cur- 
rent. Thus  when  inductive  reactance  and  condensive  reactance 
are  connected  in  series  they  oppose  and  the  reactance  of  the 
circuit  is 

X  =  XL  -  Xc  =  27T/L  -  ~~  ohms. 

In  series-parallel  circuits  the  reactance  is  a  complex  function  of 
the  resistances  and  reactances  of  the  various  branches. 
The  reactance  of  any  circuit  is 

reactive  component  of  impressed  e.m.f. 


X  = 


current 

quadrature  component  of  impressed  e.m.f. 
current 


3.  The  impedance  of  a  circuit  includes  both  the  resistance  and 
the  reactance;  it  is 


ELECTRIC  CIRCUITS  133 

4.  The  admittance  of  a  circuit  is 


Y  =  : .;  (206) 

impressed  e.m.f. 

it  is  the  reciprocal  of  the  impedance  and  thus 

Y—Z  =  VW^'  .    .   (207) 

The  admittance  has  two  components,  conductance  and  sus- 
ceptance. 

Fig.  110  shows  a  circuit  of  impedance  Z  =  \/R2  +  X2  in 
which  the  current  lags  behind  the  impressed  e.m.f.  by  an  angle 


FIG.  110. 

<j>.  The  current  I  can  be  resolved  into  two  components  in  phase 
and  in  quadrature  with  the  impressed  e.m.f.  E.  The  in-phase  or 
active  component  of  current  is 

7i  =  7  cos  0  =  |  cos  0  =  E~  =  EG,  (208) 

where 


is  the  conductance  of  the  circuit. 

The  quadrature  of  reactive  component  of  current  is 

72  =  I  sin  <t>  =  |  sin  0  =  E^2  =  EB,  (210) 

where 

X  X 

B  =  &  =  w*^ 

is  the  susceptance  of  the  circuit. 
The  total  current 


but,  by  equation  (206),  I  =  EY,  and  therefore 


Y  =       G2  +  B*.  (212) 


134r  ELECTRICAL  ENGINEERING 

5.  The  conductance  of  a  circuit  is,  from  equation  (208), 

~  _  active  component  of  current 
impressed  e.m.f. 

6.  The  susceptance  of  a  circuit  is,  from  equation  (210), 

/?  —  reactive  component  of  current  (<y-\A\ 

impressed  e.m.f. 

In  the  solution  of  series  circuits  it  is  not  necessary  to  employ 
the  terms  admittance,  conductance  and  susceptance  but  the 
solution  of  series-parallel  circuits  is  very  much  simplified  by  their 
use. 

115.  Rectangular  Coordinates. — The  simplest  method  of  deal- 
ing with  alternating-current  phenomena  is  to  express  the  e.m.fs., 
currents,  etc.,  as  the  sum  of  two  components,  one  along  a  chosen 
axis  and  the  other  perpendicular  to  it. 

i      r 

*-L^VW t 


61        I 

. 

O  ^  t  r    _  „      ..    '.  >  I  ""I 


(a)  (&1  (c) 

FIG.  111. 

In  Fig.  111(6)  which  represents  the  e.m.f.  and  current  in  a  cir- 
cuit of  impedance  Z  —  \/r2  +  x2  the  axis  is  chosen  in  the  direction 
of  the  current  and  the  e.m.f.  is  resolved  into  two  components,  e\  in 
phase  with  the  current  and  e2  in  quadrature  ahead  of  the  current. 

The  absolute  value  of  the  e.m.f.  is 


E  =  Vei2  +  622, 
and  it  leads  the  current,  which  was  chosen  as  axis  by  an  angle  0, 

where  tan  </>  =  —• 

ei 

Thus  when  the  e.m.f.  is  expressed  as  the  sum  of  two  components 
at  right  angles  both  its  magnitude  and  its  phase  are  known. 

To  distinguish  between  horizontal  and  vertical  components 
the  prefix  j  =  V—  1  is  added  to  all  vertical  components  and 
the  expression  for  the  e.m.f.  above  is 

E  =  et  +  je2.  (215) 

The  dot  is  placed  under  the  E  to  show  that  it  is  expressed  in  rec- 
tangular coordinates  and  serves  to  distinguish  it  from  its  absolute 
value. 


ELECTRIC  CIRCUITS  135 

Since     e\  =  E  cos  <£  =  Ir  and  e2  =  #  sin  <£  =  7z, 
$  =  E  cos  0  +  jE  sin  0 
=  /r  +  j/z 

(216) 


and  therefore  the  impedance  in  rectangular  coordinates  is 

Z  =  r  +  j*.  (217) 

In  Fig.  lll(c)  the  e.m.f.  is  chosen  as  axis  and  the  current  is 
behind  it  in  phase  by  an  angle  <f>  and  has  two  components  i\  in 
phase  with  the  e.m.f.  and  iz  in  quadrature  behind  it. 

The  current  may  be  written 

/  =  ii  ~  #2,  (218) 

and  this  equation  indicates  that  the  current  has  a  value 


/  =       ii*  +  iV, 

and  that  it  is  behind  the  chosen  axis  (in  this  case  the  e.m.f.)  in 
phase  by  an  angle  4>,  where 

tan  <£  =  -?•• 
*i 

Since  i\  =  I  cos  0  =  Eg  and  iz  =  I  sin  <j>  =  Eb,  where  Y  = 
Vg2  +  62  is  the  admittance  of  the  circuit,  equation  (218)  may 
be  written 

1  =  1  cos  4>  —  jl  sin  <f> 
=  Eg  -  jEb 
=  E  (g  -  jb),  (219) 

and  therefore  the  admittance  in  rectangular  coordinates  is 

Y  =  g  -  j&. 

Admittance  and  impedance  are  not  alternating  quantities  and 
their  components  are  independent  of  the  axis  of  reference,  but 
they  can  be  represented  in  rectangular  coordinates. 

A  current  multiplied  by  an  impedance  gives  an  e.m.f.  displaced 
from  it  in  phase  by  an  angle  whose  tangent  is  the  ratio  of  the  re- 
actance to  the  resistance. 

A  current  divided  by  an  admittance  gives  an  e.m.f.  displaced 
in  phase  by  an  angle  whose  tangent  is  the  ratio  of  the  susceptance 
to  the  conductance. 

Similarly  an  e.m.f.  divided  by  an  impedance  or  multiplied  by 
an  admittance  gives  a  current. 


136 


ELECTRICAL  ENGINEERING 


By  definition  a  vector  multiplied  by  j  is  turned  through  90 
degrees  in  the  counter-clockwise  direction;  when  multiplied  by 
jz  it  is  turned  through  180  degrees  and  its  sign  is  reversed. 

Therefore 

ji  =  - 1 

and 

j  =  V^n.  (220) 

Taking  this  value  for  j  alternating  quantities  expressed  in  rec- 
tangular coordinates  referred  to  a  given  axis  can  be  added,  sub- 
tracted, multiplied  and  divided  and  the  results  obtained  are 
expressed  in  rectangular  coordinates  referred  to  the  same  axis. 


I=-irfj»s 


FIG.  112. 

It  is  not  necessary  to  choose  either  the  current  in  a  circuit  or 
the  e.m.f.  as  axis  and  any  other  line  may  be  taken  as  shown  in 
Fig.  112,  but  the  e.m.f.  must  now  be  expressed  as  the  sum  of  two 
components  along  and  perpendicular  to  the  new  axis,  thus 

E  =  el  +  je*, 
and  similarly  the  current  is 

T-      •  1_        •  • 

The  e.m.f.  has  an  absolute  value 

77?  f       o"      I  o 

£j  =  •y/  Ci   ~\~  €2 
and  is  ahead  of  the  axis  by  an  angle 

i  €2  ,*  €i 

0i  =  tan"1  -  =  cos"1  -„• 
The  current  has  an  absolute  value 


and  is  ahead  of  the  axis  by  an  angle 
02  =  tan"1^  = 


ELECTRIC  CIRCUITS  137 

The  e.m.f.  leads  the  current  by  an  angle 

<f>  =  <f>i  —  02. 

The  impedance  of  the  circuit  in  rectangular  coordinates  is 
$      ei  +  je2      ei 


=  r  +  jx, 
where  the  resistance  of  the  circuit  is 

_  eiii  +  eziz  _  ejii  +  ^2^ 

ti2  +  ta2  I2 

and  the  reactance  of  the  circuit  is 

_  ezii  —  eiiz  _  ei    —  ei 


+  i22  I2 

The  admittance  of  the  circuit  is 

Y  -  i  =  ^  +  J^2  _  ^i  +  JJ2      ei  —  jet 
~  E  ~  ei+  je2  ~  ei  +  jez      ei  -  je2 
_  e\i\  +  62^2  _  -e^ii 
=  2        J 


=  g  -  jb; 
the  conductance  is 

eiii  +  ^2^2  _  eiii  +  62^ 

6i2  +  022     :  .E'2 

and  the  susceptance  is 


7   _ 

: 


The  power  factor  of  the  circuit  is 


COS  0   =   COS     $1  —   02 

=  cos  0i  cos  02  +  sin  0i  sin 


The  power  consumed  in  the  circuit  is 
P  =  EI  cos 


=  eiii  +  erf*  (221) 

and  is  the  sum  of  the  products  of  the  components  of  the  e.m.f. 
and  current  which  are  in  phase.  The  products  of  the  components 
of  the  e.m.f.  and  current  which  are  in  quadrature,  namely, 
and  62*1,  do  not  represent  power  consumed. 


138  ELECTRICAL  ENGINEERING 

The  power  may  also  be  represented  as 

75     __      T£        •       _|  • 

or 

p  =  E2g  =  eiii  +  e2i2.  (222) 

116.  Examples  in  Rectangular  Coordinates. — 1.  Find  the  cur- 
rent in  the  circuit  in  Fig.  113  in  terms  of  the  impressed  e.m.f.  and 
the  constants  of  the  circuit. 


E  =  E,  +  #2  =  /  {(n  +  r2)  +  ./(si  +  x2)(. 


t — ^AA/V — JiflflQQft> 

r2  X2 

FIG.  113. 

The  impedance  of  the  circuit  is 

Z  =  (f!  +  ra)  +  ./(zi  -f  a;,) 
and  its  absolute  value  is 


+  (xi 
and  the  absolute  value  of  the  current  is 

T7T  777 

T  £j  it/ 

Z  = 


the  power  factor  of  the  circuit  is 


The  vector  diagram  is  drawn  taking  the  current  as  the  axis. 
2.  Solve  the  circuit  in  Fig.  114. 


where 


7*1 

Qi  =  —  .,  —  ;  --  1>  and  DI  = 
2          2 


ELECTRIC  CIRCUITS 


139 


the  main  current  is 


and  its  absolute  value  is 


=  E  V(gi 


(6 


* I. 

FIG.  114. 

The  admittance  of  the  circuit  is 

Y  —  (0i  ~i~  £2)  —  j 
and  its  absolute  value  is 


Y  =  V(gi  +  g*)2  +  (bi  +  62)2. 
The  power  factor  of  the  circuit  is 


cos      = 


The  vector  diagram  is  drawn  with  the  e.m.f.  as  axis. 
IA/wA-^m?N-,  is 


FIG.  115. 

3.  Solve  the  circuit  in  Fig.  115. 

/2  —  — /  .      =  E 

r     .  $*  _    u 


63)!, 


and  therefore 


+  jh)  -  j(bt  +  6s) 


140  ELECTRICAL  ENGINEERING 

where 

__  £2  +  gs  ,  Y  _  fr2+  63 

ai 


_ 
"  (92  +  </3)2  +  (6,  +  &3)2  (02  +  </3)2  +  (6,  +  63)2 

#i  =  /ri 
and 

#  =  #!  +  #,  =  /(n  +  fl 
The  impedance  of  the  circuit  is 

Z=  ri  +  R  +jX 
and  its  absolute  value  is 


The  absolute  value  of  the  current  is 

'-I 

The  power  factor  of  the  circuit  is 


The  vector  diagram  is  drawn  with  E%  as  the  axis. 

4.  If  an  e.m.f  .  E  =  14  +  .7*38  is  impressed  on  a  circuit  and  a 
current  I  =  6  +  j  2  flows,  find  the  impedance  of  the  circuit. 
The  impedance  is 

7  -I  -  14  +  J38 
"/"       6+J2 
_14+J38       6-J2 
6  +  j2        6  -  j2 


the  resistance  of  the  circuit  is  4  ohms  and  the  inductive  reactance 
is  5  ohms. 


Fio.  116. — Constant  potential  to  constant  current. 

5.  In  Fig.  116  a  condensive  reactance  xc  is  connected  across 
the  terminals  of  a  receiver  circuit  of  variable  impedance  Z  =  r  + 
jx.  If  an  inductive  reactance  x\  =  xc  is  connected  in  the  supply 
lines  and  a  constant  e.m.f.  E\  is  impressed  on  the  terminals,  show 


ELECTRIC  CIRCUITS  141 

that  the  current  in  the  receiver  circuit  is  constant  independent 
of  the  impedance  and  power  factor. 

E    =  e.m.f.  at  terminals  of  receiver, 
-pi 

I  =  -$=  =  current  in  receiver, 
Z 

ET 

Jc  =  —  h—  =  current  in  condensive  reactance, 

-  JXc 

/!  =  I  +  Ic  =  current  in  the  line, 
J£l  =  E  +  jl&i  —  e.m.f.  impressed. 


Substituting  the  values  above 

E,  =  E  +  jx,(I  +  7C) 


!  -t) 

JZi,  since  rcc  = 


or  in  absolute  values 
and 


Since  EI  is  constant  I  is  constant  independent  of  the  impedance 
and  the  power  factor  of  the  receiver  circuit.  This  circuit  there- 
fore transforms  power  from  constant  potential  to  constant 
current. 

117.  Kirchoff's  Laws  Applied  to  Alternating-current  Circuits. 
—  Kirchoff's  two  laws  enunciated  in  Art.  90  apply  directly  to  al- 
ternating-current circuits  when  dealing  with  instantaneous  values 
of  e.m.f  s.  arid  currents;  they  also  apply  to  the  effective  values  of 
e.m.f  s.  and  currents  when  combined  in  their  proper  phase  rela- 
tions. Thus  the  vector  sum  of  all  currents  at  a  junction  is  zero; 
and  the  vector  sum  of  all  the  e.m.fs.  around  a  closed  circuit  is 
equal  to  the  sum  of  the  e.m.fs.  consumed  by  the  resistances  in 
the  circuit. 


CHAPTER  V 
COMPLEX  ALTERNATING-CURRENT  WAVES 

118.  Complex  Alternating  Waves. — The  waves  of  alternating 
current  and  voltage  dealt  with  in  the  preceding  articles  have  been 
assumed  to  be  true  sine  waves.  The  majority  of  alternating 
waves  met  with  in  electrical  engineering  approach  a  sine  form  but 
comparatively  few  are  true  sine  waves.  The  ordinary  wave  con- 
sists of  a  fundamental  sine  wave  of  the  generator  frequency  with 
certain  higher  harmonics  superimposed  on  it. 

The  maximum,  average  and  effective  values  of  such  waves  do 
not  bear  the  same  ratios  to  one  another  as  in  the  case  of  the  sine 
wave. 

The  maximum  value  is  found  as  the  maximum  ordinate  of  the 
plotted  wave  or  it  may  be  determined  mathematically  if  the  equa- 
tion of  the  wave  is  known. 

The  average  value  is  found  as  the  area  under  one-half  wave 
divided  by  the  base,  TT  radians, 

1  TT 

#avg.  =  -  I    ed0. 
^Jo 

The  effective  value  or  root-mean-square  value  is 


cW. 


The  form  factor  of  the  wave  is  defined  as  the  ratio  of  the  effect- 
ive value  to  the  average  value  and  is  denoted  by  7, 


(223) 


In  practice  the  complex  waves  are  replaced  by  "equivalent" 
sine  waves. 

142- 


COMPLEX  ALTERNATING-CURRENT  WAVES    143 

A  sine  wave  of  e.m.f.  which  is  "equivalent"  to  a  complex 
wave  is  one  which  has  the  same  effective  value  and  the  same  fre- 
quency as  its  fundamental  and  which  also  in  conjunction  with 
the  current  wave  represents  the  same  average  power. 

In  the  circuit  in  Fig.  117,  e  and  i  are  complex  waves  of  e.m.f. 
and  current.  The  voltmeter  V  indicates  the  effective  value  of  the 
e.m.f.  =  E  and  the  ammeter  A 
indicates  the  effective  value  of  the 
current  =  /.  The  wattmeter  W 
indicates  the  average  power  con- 
sumed in  the  circuit  =  P.  The 
equivalent  sine  wave  of  e.m.f.  has 
an  effective  value  E.  the  equivalent 

-  ,V  ,v     ,.  FIG.  117. 

sine  wave  of  current  has  an  effective 

value  /  and  these  two  waves  are  displaced  by  an  angle  0  which  is 

p 
such  that  P  =  El  cos  <£  or  0  =  ^7- 


119.  Examples.  —  Following  are  a  number  of  problems  dealing  with  com- 
plex alternating  waves. 

1.  If  an  e.m.f.  wave  consists  of  a  fundamental  sine  wave  of  effective  value 
100  and  a  fifth  harmonic  of  effective  value  10  passing  through  zero  in  the 
same  direction  together,  find  the  maximum  effective  and  average  values  of 
the  wave. 

The  equation  of  the  e.m.f.  wave  is 

e  =  100\/2~sin  e  +  10\/2~sin  50. 

From  the  original  assumption  the  maximum  value  is  the  sum  of  the  maxima 
of  the  fundamental  and  the  harmonic  and  its  value  is  100\/2  +  10\/2  = 
155.5.     In  any  other  case  it  would  be  found  by  solving  for  the  maximum 
value  in  the  ordinary  way  or  by  plotting  the  wave. 
The  average  value  of  the  e.m.f.  is 


#avg.  =  i"  ede  =  I*  (100  V2  sin  6  +  10  \/2  in  50)d0. 


ioV2T  _ 

7T         L 


10  cos  ,  _ 


-I- 

Jo 


0  4i 


91.9. 


The  effective  value  is 


sin  6 


50)2d0 


=  l°V2^f*  (100  sin2  0+20  sin  0  sin  50  +  sin2  50)d0 

=  10\/2^J0'rj—^(l-cos20)+y  (cos40-cos60)+K(l-cosl00)  Id0 


144 


ELECTRICAL  ENGINEERING 


+  %TT    =  10V2V505  =  100.5. 


The  form  factor  of  the  wave  is 


Ees.        100.5 


E. 


avg. 


=  1.093. 


2.  If  the  e.m.f.  of  the  last  example  is  impressed  on  the  terminals  of  the 
circuit  CD  in  Fig.  118(o)  determine  the  current  flowing.  The  fundamental 
frequency  is  50  cycles  per  second. 

The  resistance  of  the  circuit  is  1  ohm,  the  inductive  reactance  is  2  ohms 
at  50  cycles  and  10  ohms  at  250  cycles  and  the  condensive  reactance  is  50 
ohms  at  50  cycles  and  10  ohms  at  250  cycles. 

The  current  wave  is  made  up  of  a  fundamental  of    /-.    ,      7;  --  9^  =  — 

=  2.08  amp.  leading  the  fundamental  e.m.f  .  by  nearly  90  degrees  and  a  fifth 
10 

harmonic  of  =  1 


The  circuit  is  resonant  for  the  fifth  harmonic  of  e.m.f.  at  250  cycles  and  so 
the  fifth  harmonic  current  is  very  much  exaggerated.  The  current  is  plotted 
in  Fig.  118(6). 


X=  2  Ohma  at  60  Cycles 
X  =  10  Ohms  at  250  Cycles 


c=  60  Ohms  at  50  Cycles 
Xc=10  Ohms  at  250  Cycles 


FIG.  118. 

If  the  e.m.f.  wave  maintains  its  shape  and  value  while  the  frequency  of  the 
fundamental  is  varied  from  0  to  500  cycles  per  second,  plot  the  effective 
values  of  the  current  on  a  frequency  base. 

In  the  table  below  are  given  the  effective  values  of  the  fundamental  and 
fifth  harmonic  currents  at  various  frequencies  and  also  the  effective  values 
of  the  resultant  current. 

The  values  and  phase  relations  of  the  fundamental  and  fifth  harmonic 
currents  are  found  by  treating  the  fundamental  and  fifth  harmonic  e.m.fs. 
separately.  The  equation  of  the  resultant  current  can  then  be  written  down 
and  the  effective  value  found  as  in  example  1. 

At  50  cycles  the  equation  of  the  current  wave  is  i  =  2.08-\/2  sin  (0  +  90) 
+  lQ\/2  sin  56  and  its  effective  value  is  10.21  amp. 


COMPLEX  ALTERNATING-CURRENT  WAVES    145 


/ 

5/ 

Fundamental 
current 

Fifth  harmonic 
current 

Resultant 
current 

0 

0 

0.00 

0.00 

0.00 

50 

250 

2.08 

10.00 

.       10.21 

100 

500 

4.76 

6.60 

8.12 

200 

1,000 

21.70 

0.27 

21.70 

250 

1,250 

100.00 

0.21 

100.00 

300 

1,500 

26.70 

0.17 

26.70 

400 

2,000 

10.20 

0.13 

10.20 

500 

2,500 

6.70 

0.10 

6.70 

GO 


040 


20 


"0  100  200  300  400  500 

Freauency  of  the  Fundamental  In  Cycles  per  Sec. 

FIG.  119. 

These  values  of  current  are  plotted  on  a  frequency  base  in  Fig.  119.  There 
are  two  peaks  on  the  curve,  first  at  50  cycles  where  the  circuit  becomes 
resonant  for  the  fifth  harmonic  and  again  at  250  cycles  where  the  circuit 
becomes  resonant  for  the  fundamental. 

^Impressed  E.M.r. 

Fundamental 


A      I          (1)                 (2)                 (3 

Xc  =10  Ohms 
at  50  Cycles 

E               >Rs 

!„     X=100hm.T 
g  at  60  Cycles  1X 

§  X  =  600hms 
*p  at  250  Cycles 

at  250  Cycles 

•          ..     (6) 

: 

8I     4- 

0 

' 
Fundam< 
and 

10 

>     »  E 

0 

fifth 
Curr 
J 

(«) 

Harmonic 

juts  and 
...M.i'. 

10    I  -I 

10 

(d)        i 

:ntal  Currents 
E.M..F. 

FIG.  120. 

3.  If  the  same  e.m.f.  wave  is  impressed  on  the  parallel  circuit  AB  in  Fig. 
120,  find  the  current  in  the  three  branches  at  a  frequency  of  50  cycles  per 
second. 

The  resistance  of  branch  (1)  is  10  ohms;  the  reactance  of  (2)  is  10  ohms  at 
10 


146  ELECTRICAL  ENGINEERING 

%». 

50  cycles  and  50  ohms  at  250  cycles;  the  condensive  reactance  of  (3)  is  10 
ohms  at  50  cycles  and  2  ohms  at  250  cycles.  ^~ 

The  current  /i  in  (1)  consists  of  a  fundamental  of  -777  =  10  amp.  and  a 

10 

fifth  harmonic  of  y^  =  1  amp.  in  phase  with  their  respective  e.m.fs.      The 

equation  of  the  current  is 

ii  =  10\/2  sin  0  +  \/2  sin  50. 

The  wave  shape  is  the  same  as  that  of  the  impressed  e.m.f.  Resistance, 
therefore,  does  not  affect  the  wave  shape.  10~ 

The  current  72  in  (2)  consists  of  a  fundamental  of  -^  r-  =  10  amp.  and  a 

10 

fifth  harmonic  of  ^  =  0.2  amp.  in  quadrature  behind  the  e.m.fs.  pro- 
ducing them.  The  equation  of  the  current  is 

it  =  10\/2  sin  (8  -  90)  +  0.2  X  \/2  sin  (56  -  90) 

The  fifth  harmonic  is  not  nearly  so  prominent  as  in  the  resistance  circuit. 
Inductive  reactance  thus  tends  to  eliminate  the  harmonics  in  irregular  waves 
and  makes  them  more  sinusoidal  in  form.  in~ 

The  current  73  in  (3)  consists  of  a  fundamental  of  ^TTT  =  10  amp.  and  a 

10 

fifth  harmonic  -^   =  5  amp.  in  quadrature  ahead  of  the  e.m.fs.  producing 

them.     The  equation  of  the  current  is 

iz  =  lO-s/2  sin  (6  +  90)  +  5\/2  sin  (50  -f  90). 

The  fifth  harmonic  is  much  more  prominent  in  the  capacity  circuit  than  in 
either  of  the  others.  Capacity,  therefore,  tends  to  exaggerate  the  har- 
monics in  a  peaked  wave. 

The  main  current  I  consists  of  a  fundamental  and  a  fifth  harmonic.  The 
fundamental  is  the  resultant  of  the  fundamental  currents  in  the  three 
branches  and  from  Fig.  120 (d)  is  found  to  be  10  amp.  in  phase  with  the 
fundamental  e.m.f.  The  fifth  harmonic  is  the  resultant  of  the  fifth  harmonic 
currents  in  the  three  branches  and  from  Fig.  120 (e)  it  is  found  to  be  4.8  amp., 
leading  the  fifth  harmonic  e.m.f.  by  approximately  a  quarter  of  a  cycle.  The 
equation  of  the  resultant  current  is 

i  =  10\/2  sin  6  +  4.8  X  \/2  sin  (50  -f-  90). 

This  current  is  shown  in  Fig.  120  (c). 

20.  Analysis  of  Alternating  Waves. — The  majority  of  alter- 
nating-current problems  are  solved  on  the  assumption  that  the 
waves  of  e.m.f.  and  current  are  sine  waves  or  such  that  they 
can  be  represented  by  sine  waves  with  sufficient  accuracy.  In 
certain  cases,  however,  it  is  necessary  to  know  the  values  of  the 
more  important  harmonics  in  order  to  understand  the  phenomena 
completely. 

The  most  general  expression  for  a  univalent  periodic  function 
may  be  given  in  the  form  of  an  infinite  trigonometric  series  or 
Fourier  series, 
y  =  do  +  «i  cos  0  -f-  02  cos  26  -\ — \-  an  cos  nB  -f-  61 

sin  B  +  62  sin  26  -f  +  bn  sin  nO     (224) 


COMPLEX  ALTERNATING-CURRENT  WAVES    147 

where  ao,  a\,  a2,  61,  62,  etc.,  are  constants;  or  combining  the  sine 
and  cosine  terms  it  may  be  expressed  as 

y  =  a0  -f  Ci  sin  (0  +  <£i)  +  C2  sin  (20  +  <£2)  +  + 

cn  sin  (n0  +  </>„)     (225) 
where 

,— 2 -  = On t  = 6n 

tan  <f>n  =  ^-  (226) 

If  the  ordinates  of  such  a  function  for  one  complete  cycle  of 
the  fundamental,  are  given  in  form  of  a  curve  as  obtained  from 
an  oscillograph  record  or  tabulated  as  on  page  149,  the  constants 
an  and  bn  for  any  harmonic  can  be  found  as  explained  below. 

To  find  the  coefficient  ao  integrate  equation  (224),  from  0  to  2ir. 

r/»2,r 
ydd  =    I     (ao  +  a\  cos  0  +  a2  cos  20  -f  +  +  an   cos 
Jo 
nd  +  &i  sin  0  +  &2  sin  20  +  +  +  bn  sin  n0)d0  =  a0[0]oir  =  27ra0, 

since  all  the  other  terms  vanish  between  these  limits. 

Thus  a0  =  o~  =  average  value  of  the  ordinate  between  0  and  2ir. 

The  coefficient  an  is  found  by  multiplying  all  the  terms  in  equa- 
tion (224)  by  cos  n8  and  integrating  from  0  to  2ir.     This  gives 

An  =   I     y  cos  nddd  =    I      (ao  cos  nQ  +  ai  cos  nd-  cos  0  H — | — h 
Jo  Jo 

an  cos2n0  -f  &i  cos  n&  sin  0  H — h  +  bn  cos  nd  sin  n0)d0 

I  a0  cos  nd  +  -jjcbft'Oi  -  1)0  +  cos  (n  +  1)0}  +  -h  + 
/> 

^(1  +  cos  2n0)  +  Trfsin  (n  +  1)0  -  sin  (n  -  1)0}  +  +  + 


I 


and  oM»\       o  /  a\27r 

an  =     \2^/  =      avg>  ^  cos  n  ^o 

bn  is  found  by  multiplying  equation  (224)  by  sin  n0  and  integrat- 
ing as  before. 

£2*  £2* 

Bn  =    I     y  sin  n0d0  =    I     (a0  sin  nB  +  0,1  sin  n0  •  cos  0  +  +  + 
Jo  Jo 

an  sin  n0  cos  nd  +  61  sin  n0-'sin  0  +  +  +  bn  sin2  n6)dO 


148  ELECTRICAL  ENGINEERING 

=    I     [a0  sin  nS  +  ^{sin  (n  -f  1)0  +  sin  (n  -  1)0}  +  +  + 
^(sin2n0)  +  ~  {cos  (n   -  1)0  -  cos  (n  +  1)0}  +  +  + 


^(1  —  cos  2n0)Jd0 


and 

6n  =  2\2W  =  2  avg*  ^  sin  n^o* 
Applying  these  results, 
a0  =  avg.  (y)l' 

ai  =  2  avg.  (y  cos  0)5T         61  =  2  avg.  (y  sin  0)^ 
a2  =  2  avg.  (y  cos  26)1'       62  =  2  avg.  (y  sin  20)ox 
an  =  2  avg.  (y  cos  n0)o7r       6n  =  2  avg.  (#  sin  710)^ 

Practically  all  of  the  waves  met  with  in  electrical  engineering 
are  symmetrical  waves  having  the  two  half  waves  of  the  same 
shape  but  opposite  in  sign.  The  values  of  the  ordinates  from 
180  to  360  degrees  are  the  same  as  those  from  0  to  180  degrees 
but  their  signs  are  different. 

In  symmetrical  waves  only  those  harmonics  can  exist  which 
reverse  in  sign  when  0  is  increased  by  180  degrees.  Therefore, 
all  the  even  harmonics  are  absent  and  there  is  no  constant  term. 
The  general  expression  for  a  symmetrical  alternating  wave 
may  be  put  in  the  form 
y  =  a\  cos  0  +  ^3  cos  30  +  a5  cos  50  +  +  + 

61  sin  0  +  63  sin  30  ++     (227) 
or  combining  the  sine  and  cosine  terms 
y  =  ci  sin  (0  +  </>i)  +  c3  sin  (30  +  #3)  +  c5  sin 

(30  +  0B)     (228) 

where  the  values  ci,  c3,  <j>i,  $3,  etc.,  are  found  as  indicated  in 
equation  226. 

In  order  to  analyze  such  a  wave  into  its  component  harmonics 
it  is  necessary  to  know  the  values  of  the  ordinates  for  one-half 
wave  only  and  the  constants  are  determined  from  the  following 
equations, 

ai  =  2  avg.  (y  cos  6)'  bi  =  2  avg.  (y  sin  6)' 

a3  =  2  avg.  (y  cos  30)^  63  =  2  avg.  (y  sin  30)* 

a5  =  2  avg  (y  cos  50)*  65  =  2  avg.  (y  sin  50)* 


COMPLEX  ALTERNATING-CURRENT  WAVES    149 

121.  Example  of  Analysis. — One-half  of  an  alternating-current 
wave  is  shown  as  curve  1,  Fig.  121,  and  it  may  be  represented  by 
the  expression, 

y  =  ai  cos  0  +  as  cos  30  +  a5  cos  50  +  +  + 

61  sin  6  +  63  sin  30  +  68  sin  50  +  + 
or  by 

y  =  Ci  sin  (0  -f  <£i)  +  c3  sin  (30  +  03)  +  c5  sin  (50  +  05)  +  higher 

harmonics.  - 

The  values  required  to  determine  the  fundamental,  and  third 
and  fifth  harmonics  are  tabulated  below. 


1 

1 

<£> 

* 

§ 

fl 

£ 

3 

CO 

CO 

1C 

§ 

5 

S 

0 

a 

8 

•53 

a 

8 

.2 

a 

o 

c 

%» 

A 

a. 

6 

•a 

Si 

s> 

°5 

a 

•a 

Q 

£ 

0 

0.0 

0.0 

,000 

0.000 

0.00 

0.00 

1.000 

0.000 

0.00 

0.00     1.000 

O.OCO!      0.00 

0.00 

10 

5.0 

25.0 

0.985 

0.174 

4.85     0.871    0.866 

0.500 

4.33 

2.50    0.643 

0.766 

3.22 

3.84 

20 

8.2 

67.2 

0.940 

0.342 

7.70    2.80    0.500 

0.866 

4.10 

7.10-0.174 

0.985 

-  1.43 

8.07 

30 
40 

9.7 
11.0 

94.0 
121.0 

0.8660.500 
0.7660.643 

8.40 
8.42 

4.85J    0.000 
7.07i-0.500 

1.000 
0.866 

0.00 
-  5.50 

9.70-0.866     0.500 
9.53-0.940-0.342 

-  8.40 
-10.34 

4.85 
-  3.76 

50 

12.0 

144.0 

0.6430.766 

7.71     9.19-0.866 

0.500 

-10.40 

6.00-0.342'-0.940 

-  4.10-11.28 

60 

13.7 

187.7 

0.5000.866 

6.85  11.86-1.000 

0.000 

-13.70 

O.OOi    0.500!-0.866 

6.85-11.85 

70 

15.1 

228.0 

0.3420.940 

5.16 

14.15-0.866 

-0.500 

-13.08 

-  7.55     0.985-0.174 

14.87 

-  2.62 

80 

17.7 

313.3 

0.1740.985 

3.70 

17.40-0.500 

-0.866 

-  8.85 

-15.32     0.766 

0.643 

13.56 

11.37 

90 

25.7 

660.5 

0.000 

1.000 

0.00 

25.70 

0.000 

-1.000 

0.00 

-25.70    0.000 

1.000      0.00 

25.70 

100 

30.5 

930.2 

-0.174'0.985 

-    5.28 

30.00 

0.500 

-0.866 

15.25 

-26.40-0.766 

0.643-23.40 

19.60 

110 

38.5 

1482.2 

-0.3420.940-  13.30 

36.00 

0.866 

-Of500 

33.35 

-19.25-0.985 

-0.174-37.90 

-  6.70 

120 

44.0 

1936.0 

-0.5000.866-  22.00 

38.10 

1.000 

0.000 

44.00 

0.00-0.500-0.866 

-22.00-38.20 

130 

46.0 

2116.0 

-0.6430.766-  28.60 

35.30 

0.866 

0.500 

39.85 

23.00     0.342;  -0.940 

15.70-43.20 

140 

40.8 

1664.6 

-0.7660.643-  31.30 

26.20 

0.500 

0.866 

20.40 

35.30     0.940-0.342 

38.35 

-13.83 

150 

30.0 

900.0 

-0.866 

0.500-  26.00 

15.00 

0.000 

1.000 

0.00 

30.00     0.866 

0.500 

25.98 

15.00 

165 

19.0 

361.0 

-0.9400.342-  17.87 

6.50 

-0.500 

0.866 

-  9.50 

16.46     0.174 

0.985 

3.30 

18.70 

170 

10.0 

100.0 

-0.985 

0.174-    9.85 

1.74 

-0.866 

0.500 

-  8.66 

5.00-0.643 

0.766 

-  6.43 

7.66 

Total  376.  9  11330.7 

-101.41282.73 



91.59 

50.37  

1 

7.83 

-16.65 

ai  =  2  avg.  (y  cos  0)' 

03  =  2  avg.  (y  cos  30  J            i  a&  =  2  avg.  (y  cos  50  j  * 

/  — 

101.41v 

11.27 

-=  2 

/91.5£ 

)  -  10.17 

=  2 

/7.83> 
\  18  ; 

-  0.87 

"  "V       18       )  ~ 

V    18 

61  =  2  avg.  (y  sin  0  J 

6s  =  2  avg.  (y  sin  30)' 

65  -  2  avg.  (y  sin  50)' 

_      /282.73s 

=  31.41 

=  2 

/50.3S 

)  =  5.596 

=  2 

/-16.65N 

-1.85 

"A    18     / 

V   18 

^       18 

) 

ci   =  V  a\ 

2  +  612  =  33.3 

cs  -  Vaz2  +  W  =  11.6 

C5    =    > 

/as2  +  6s2  -  2.05 

sin  0i 

ai        -  11.27 

1.338 

sin03=^  = 

10.17 

0.877 

sin  05 

as 

0.87 

=  0.434 

ci            33.3 

11.6 

es 

2.05 

cos  0    =  b 

31.41 

CU4 

63       5.596 

65 

-1.85 

a        33.3   ~ 

°S           ca        11.6 

COS  05 

C5 

2.05 

=  -  0.902 

tan0!-- 

i        -11.27 

Q 

.359 

as 

10.17 

=  1.82 

tan  06=  r-S  «= 

06 

0.87 

i         31.41 

5.596 

-1.85 

--0.47 

0!    -     - 

-  19°  45' 

03  -  61°  10' 

08  -  154°  49' 

150 


ELECTRICAL  ENGINEERING 


The  fundamental  is  yi  =  Ci  sin  (0  ~f  0i)  =  33.3  sin  (0  —  19° 45'), 
Curve  2,  Fig.  121;  the  third  harmonic  is  3/3  =  Cs  sin  (30  +#3) 
=  11.6  sin  (30  +  61°  10'),  Curve  3,  Fig.  121;  the  fifth  harmonic 
is  2/5  =  c5  sin  (50  +  06)  =  2.05  sin  (50  +  154°  49'),  Curve  4,  Fig. 


25   50   75   100   125   150  175   200 
FIG.  121. 


121;  and  the  complete  expression  for  the  alternating-current 
wave  is  y  =  33.3  sin  (0  -  19°  45')  +  11.6  sin  (30  +  61°  10') 
+  2.05  sin  (50  +  154°  49')  +  higher  harmonics  which  are 
negligible. 


no 


100 


40  50 


FlG.   122. 


The  maximum  ordinate  taken  from  the  plotted  curve  is 

2/ma*.  =  46; 

The  average  ordinate  is  found  from  column  (1) 

376.9 


2/avg. 


18 


=  20.9; 


COMPLEX  ALTERNATING-CURRENT  WAVES    151 
the  effective  value  is  obtained  from  column  (3) 


_      11,330.7 

2/eff.    - 


and  the  form  factor  is 


The  effective  value  may  also  be  found  by  plotting  the  values  of 
y  on  polar  coordinate  paper  as  in  Fig.  122  and  measuring  the  area 
of  the  curve  Ap  by  means  of  a  planimeter. 

The  area  of  a  polar  curve  is 


Ap  = 
and  therefore  the  effective  value  is 


o 


The  wave  analyzed  above  was  taken  from  an  oscillograph 
record  of  the  exciting  current  of  a  transformer. 


CHAPTER  VI 
POLYPHASE  ALTERNATING-CURRENT  CIRCUITS 

122.  Polyphase  Alternating-current  Circuits. — In  Art.  102 
the  generation  of  an  alternating  e.m.f.  was  discussed.  Fig.  123 
represents  a  simple  single-phase  alternator.  The  field  poles  are 
excited  by  direct  current  and  an  alternating  e.m.f.  is  produced  in 
the  winding  which  is  carried  on  the  armature.  It  does  not  make 


Sl  Armature  Winding  fl 


FIG.  123. — Single-phase  alternator,  revolving  field  type. 


any  difference  whether  the  poles  are  fixed  and  the  armature  wind- 
ing moves  or  the  winding  is  fixed  and  the  poles  move. 

The  winding  starts  at  Si  and  ends  at  f\  and  the  equation  of  the 

e.m.f.  between  s\  and  /i  is 

-pi 

ei  =  Em  sin  6,  effective  value  =  — %  =  E 


A  BC  D 


FIG.  124. — Two-phase  winding. 

The  positive  direction  of  e\  is  from  Si  to  /i.  It  is  here  assumed 
that  the  flux  in  the  air  gap  is  so  distributed  that  the  e.m.f.  is  a 
sine  wave. 

If  a  second  winding  is  placed  on  the  same  armature,  Fig.  124, 
but  displaced  90  electrical  degrees  from  the  first,  it  will  have  an 

152 


POLYPHASE  ALTERNATING-CURRENT  CIRCUITS    153 


e.m.f.  generated  in  it  of  the  same  value  and  wave  form  as  the 
first  but  displaced  from  it  in  phase  by  90  degrees  (Figs.  125  and 
126). 
Phase  2  starts  at  s2  and  ends  at/2  and  the  e.m.f.  generated  in  it 

is 

•tm 


02  =  Em  sin  (6  —  90),  effective  value 
The  machine  is  called  a  two-phase  alternator. 


\/2 


E  = 


ElectricalNDegrees 


FIG.  125. — E.m.f.  waves  of  a  two- 
phase  alternator. 


FIG.  126. 


The  two  windings  are  usually  entirely  separate  and  their  ends 
are  brought  out  to  four  terminals  ABCD. 

If  any  two  terminals  as  B  and  C  are  joined,  the  e.m.f.  between 
A  and  D  is 

eAD  =  ei  —  e2  =  Em  sin  0  —  Em  sin  (6  —  90) 

=  Em  (sin  0  +  cos  0)  =  V2#m  sin(0  +  45); 

it  leads  e\  in  phase  by  45  degrees,  its  maximum  value  is   \2-6Jm  and 
its  effective  value  is  EAD  = 


E 


V2E.     This  value 

A     B      C     D      F 


can  also  be  obtained  by  subtract- 
ing the  two  vectors  as  shown  in 
Fig.  126. 

The  middle  points  mi  and  ra2 
of  the  two  windings  are  some- 
times connected  together  and  a 
fifth  terminal  F  used  as  shown 
in  Fig.  127.  The  common  term- 
inal F  is  called  the  neutral  point 

of  the  winding  and  may  be  connected  to  earth.     The  e.m.f.  from 
each  of  the  other  terminals  to  F  is  the  same, 

_  E 


/I 

°mi,m%    f 

(((OTTOCOTIT)   2 

0(footfo(ftrv- 

8l 

g 

FIG.  127. 


154 


ELECTRICAL  ENGINEERING 


The  four  e.m.fs.  EAC,  ECB,  EBD  and  EDA  are  equal,  since  each  is 

7? 
the  vector  difference  of  two  e.m.fs.  of  effective  value  -~  at  right 

angles  to  one  another,  and  these  four  e.m.fs.  are  also  at  right 
angles  to  one  another  and  form  a  four-phase  or  quarter-phase 
system. 

The  effective  value  of  each  of  the  four  e.m.fs.  is 

E 


123.  Three-phase  Circuits. — If  three  similar  windings  are 
placed  on  the  same  alternator  armature  displaced  120  electrical 
degrees  from  one  another,  Fig.  128,  and  the  ends  of  the  windings 


FIG.   128. — Three-phase  winding. 

are  brought  out  to  terminals,  the  machine  is  a  three-phase  alter- 
nator.    The  e.m.fs.  generated  in  the  three  windings  are  dis- 
placed 120  degrees.  ™ 
The  e.m.f.  in  phase  1  is  e\  =  Em  sin  0,  effective  value  E  =  ~~^=; 

the  e.m.f.  in  phase.  2  is  e2  =  Em  sin  (0  —  120),  effective  value 

E 

E  =  — p;  and  the  e.m.f.  in  phase  3  is  e3  =  Em  sin  (0  —  240), 


effective  value  E  = 


ABC 


FIG.   129. — Delta  connection. 


The  windings  may  be  interconnected  in  two  ways.  (1)  Join 
/i  to  $2,  /2  to  ss  and  /3  to  Si,  and  connect  the  three  junctions  to 
the  terminals  A,  B  and  C  (Fig.  129)  This  is  called  the  "delta" 
connection  or  ring  connection  and  is  represented  by  A. 


POLYPHASE  ALTERNATING-CURRENT  CIRCUITS    155 


The  resultant  e.m.f.  around  the  closed  circuit  at  any  instant  is 

+  e2  +  <?3  =  Em  sin  6  +  Em  sin  (0  -  120)  +  Em  sin  (0  -  240) 
=  Em  (sin  0  +  sin  0  cos  120  -  cos  6  sin  120 
+  sin  6  cos  240  -  cos  0  sin  240) 


Em  (sin  0  -  2  sin  0 ^-  cos  0 


sin  0 


cos  0     = 


).„. 


"2.  If  the  ends  «i,  $2  and  §3  are  connected  together  and  the  ends 
/i,  /2  and/s  are  joined  to  the  three  terminals  A,  B  and  C,  Fig.  130, 
the  windings  are  connected  Y  or  "star." 


FIG.   130. — Star  or  "  Y"  connection. 

The  e.m.f.  between  A  and  B  is 

eAB  =  ei  -  62  =  Em  sin  0  -  Em  sin  (0  -  120) 
=  Em(sm  0  -  sin  0-cos  120°  +  cos  0-sin  120) 

sin  0  H — pr-  cos  0 


=  #m    sin  0  + 


m(  -^-  sin  0  +  M  cos  0 ) 

.  (0  -f  30); 
its  maximum  value  is  \/3  -^m  and  its  effective  value  is 


(230) 


Similarly  the  e.m.f.  between  B  and  C  is 

eBC  =  e2  -  63  =  Em  {sin  (0  -  120)  -  sin  (0  -  240) ) 

=  V3Em  sin  (0  -  90)  (231) 

of  maximum  value  V3  Em  and  effective  value  V3#. 
The  e.m.f.  between  C  and  A  is 

eCA  =  es  -  ei  =  Em  {sin  (0  -  240)  -  sin  0} 

=  VZEm  sin  (0  -  210)  (232) 

of  maximum  value  V3Em  and  effective  value 


156 


ELECTRICAL  ENGINEERING 


Thus  the  three  e.m.fs.  between  terminals  are  equal  to  one 
another  and  are  displaced  in  phase  by  120  degrees. 

A  fourth  terminal  is  usually  connected  to  the  neutral  point  0 
and  it  may  be  grounded. 

124.  Electromotive  Forces,  Currents  and  Power  in  Three- 
phase  Circuits.  —  Fig.  131  shows  a  delta-connected  three-phase 
system. 


ET 


FIG.  131. 

The  e.m.f.  in  phase  1  is 

€1  =  Em  sin  6,  effective  value  E\  = 
the  e.m.f.  in  phase  2  is 

e2  =  Em  sin  (0  -  120),  effective  value  E2  = 
the  e.m.f.  in  phase  3  is 

e3  =  Em  sin  (6  —  240),  effective  value  E3  = 


Em 

V2' 

E^_ 

V2* 


The  effective  values  of  the  e.m.fs.  in  the  three  phases  are  equal 
and  are  the  terminal  e.m.fs.  of  the  alternator  or  the  e.m.fs. 
between  lines, 

Tjl 

Et  =  EI  =  EZ  =  ES  = 


Assume  that  the  loads  on  the  three  phases  are  balanced,  that 
is,  that  the  currents  have  the  same  effective  value  and  are  dis- 
placed from  their  respective  e.m.fs.  by  the  same  angle  </>. 

The  current  in  phase  1  is 

i\  =  Im  sin  (B  —  </>),  effective  value  7i  = 
the  current  in  phase  2  is 

iz  =  Im  sin  (B  —  $  —  120),  effective  value  1  2  = 
the  current  in  phase  3  is 

?3  =  Im  sin  (B  —  <£  —  240),  effective  value  J3  =  —=' 


POLYPHASE  ALTERNATING-CURRENT  CIRCUITS    157 


The  effective  value  of  the  currents  in  the  three  phases  is  the  same 
and  is  7  = 


The  current  in  line  A  is 

iA  =  u  -  *2  =  7»_sin  (0  -  <t>)  -  Im  sin  (0  -  4>  -  120) 

=  V37msin  (0  -  <f>  +  30),  (233) 

and  its  effective  value  is 

IA   =   V3   ^%=   \/3  7. 

The  current  in  line  B  is 

iB  =  i2  -  is  =  Im  sin  (0  -  4>  -  120)  -  Im  sin  (B  -  <f>  -  240) 
=  A/37™  sin  (0-0-90),  (234) 

and  its  effective  value  is 

IB  =  V3I- 

The  current  in  line  C  is 
ic  =  is  —  ii  =  Im  sin  (6  -  <£  -  240)  -  Im  sin  (0  -  <£) 

=  V3Jm  sin  ((9  -  0  -  210),  (235) 

and  its  effective  value  is 

Ic  =  V3/. 


Thus  the  currents  in  the  three  lines  are  equal  in  magnitude  and 
are  120  degrees  out  of  phase  with  one  another.  If  the  effective 
value  of  the  current  in  each  of  the  lines  is  represented  by  Ii  then 

Ii  =  I  A  =  IB  =  Ic  =  V3I. 

The  effective  values  of  all  these  quantities  are  shown  in  the  vector 
diagram  in  Fig.  132. 


158 


ELECTRICAL  ENGINEERING 


The  power  supplied  by  the  alternator  is 

P  =  Eili  cos  <£  +  Ezlz  cos  0  +  E3Is  cos  <£ 

=  3EI  cos  $  (236) 

=  \/ZEtIi  cos  0  (237) 

and  is  equal  to  \/3  times  the  product  of  the  terminal  e.m.f.,  the 
line  current  and  the  power  factor. 

If  the  system  is  not  balanced,  that  is,  if  either  the  currents  or 
the  power  factors  in  the  three  phases  differ  from  one  another  the 
line  currents  will  not  be  equal  and  they  will  not  be  displaced  in 
phase  by  120  degrees. 

If  the  current  in  phase  1  is 

i\  =  Imi  sin  (0  —  fa) 
and  the  current  in  phase  2  is 

*2  =  /™2sin  (0  -  02  -  120) 
the  current  in  line  A  is 

=  Imi  sin  (0  -  fa)  -  Im2  sin  (0  -  fa  -  120).      (238) 
Fig.  133  shows  a  star-  or  F-connected  three-phase  system 


FIG.  133. 

Using  the  same  notation  as  before  and  taking  the  results  ob- 
tained in  Art.  123, 

the  e.m.f.  between  lines  A  and  B  is 

eAB  =  ei-e2  =  \/liEm  sin  (0+30),  effective  value  EAB= 

the  e.m.f.  between  lines  B  and  C  is 

eBc  =  e2-e3  =  \/3Em  sin  (0-90),  effective  value  EBC  = 

the  e.m.f.  between  lines  C  and  A  is 

Sin  (0-210),  effective  value  ECA  = 


POLYPHASE  ALTERNATING-CURRENT  CIRCUITS    159 

the  current  in  line  A  is 

iA  =  i\  =  I-m  sin  (B  —  0),  effective  value  IA  =  —7=  =  //; 
the  current  in  line  B  is 

Im 

IB  =  ii  =  Im  sin  (B  —  <f>  —  120),  effective  value  IB  =  ~~7^  =  Ii', 
the  current  in  line  C  is 

ic  =  is  =  Im  sin  (6  —  <f>  —  240),  effective  value  Ic  =  ~^ 
Fig.  134  is  a  vector  diagram  for  this  circuit. 

EA8 


•® 


FIG.   134. 


125.  Measurement  of  Power    in  Polyphase    Circuits. — The 

power  in  two-phase  circuits   (Fig.   135)   can  be  measured  by- 
connecting  wattmeters  in  the  two  phases  and  taking  the  sum  of 


Phase  1 


Phase  1 


Load 


FIG.  135. — Measurement  of  power  in  two-phase  circuits. 

the  readings.     If  the  two  phases  are  similarly  loaded  a  single 
meter  will  suffice  if  its  reading  is  doubled. 

In  the  three-phase  circuits  (Fig.  136)  the  total  power  is  given 
by  the  sum  of  the  two  wattmeter  readings,  if  they  are  connected, 


160 


ELECTRICAL  ENGINEERING  - 


as  shown  at  a,  with  their  current  coils  in  two  of  the  lines  and  their 
voltage  coils  connected  across  from  these  lines  to  the  third  line. 


Terminal  Voltage 
Line  Current 


FIG.  136. — Power  in  balanced  three-phase  circuits. 

Consider  first  the  case  where  the  loads  on  the  three  phases 
are  balanced.  Referring  to  the  vector  diagram,  Fig.  137,  for 
the  Y  —  connected,  circuit, 

Wi  =  EABIA  cos  (30  +  0)  -  \/3EI  cos  (30  +  0). 
W2  =  ECBIC  cos  (30  -  0)  =  \/3EI  cos  (30  -  <#>). 


FIG.  137. 


Adding, 

Wi  +  TF2  = 


{cos  (30  +  0)  +  cos  (30  -  0)} 
{(cos  30  cos  0  —  sin  30  sin  <£)  +  (cos  30 
cos  0  +  sin  30  sin  $) } 
=  \fSEI  (2  cos  30  cos  0)  =3#/cos<£  =  \/3#Jj  cos  <£ 

and  this  is  the  total  power  in  the  circuit. 

The  power  factor  of  the  circuit  can  also  be  found  from  the  two 
wattmeter  readings. 
Subtracting, 

W 2  -  Wi  =  \/3EI  (2  sin  30  sin  0)  =  \/3EI  sin  0. 


and 


W  2  - 


sn 


_ 

:  :        tan 


POLYPHASE  ALTERNATING-CURRENT  CIRCUITS    161 


Gl- 


and 


tan  <f>  =  V3 


W 2  - 


COS         = 


(239) 
(240) 


When  the  power  factor  is  50  per  cent.,  $  =  60  degrees  and 
the  reading  of  wattmeter  Wi  becomes  zero, 

Wi  =  EABIA  cos  (30  +  60)  =  0. 

For  lower  power  factors  the  reading  of  W\  would  be  negative 
and  the  total  power  in  the  circuit  is  given  by  the  difference  of 
the  readings. 

In  some  cases  it  may  be  difficult  to  tell  whether  the  readings 
should  be  added  or  subtracted.  The  curve  in  Fig.  138  makes  it 
possible  to  determine  the  sign  of  Wi.  It  shows  the  relation 
between  power  factor  and  the  ratio  of  the  readings  of  the  two 


meters,  cos  cj>  vs. 


w. 


-1.0 


-0.5  0    ... 

Ratio     n  »        * 
W2 

FIG.  138. — Watt-ratio  curve. 


0.5 


1. 


The  equation  of  the  curve  may  be  obtained  from  equation  (240)  . 


Cos  0 


1 


V (W 2  -I-  Wtf  +  3(TF 2  -  W i)2 


n 


162 


ELECTRICAL  ENGINEERING 


Above  50  per  cent,  power  factor  W\  is  positive  and  v^-  is  positive. 

Wi 
At  50  per  cent,  power  factor  Wi  and  wr    become  zero.     Below 

W  i 

50  per  cent,  power  factor  W\  and  ™r-  are  negative.     If  in  any 

W  2 

case  it  is  doubtful  whether  W\  is  positive  or  negative,  try  it 
both  ways  and  calculate  the  corresponding  value  of  cos  $  from 
equation  (241)  and  then  referring  to  Fig.  138  see  which  assump- 
tion agrees  with  the  watt-ratio  curve. 

If  the  current  and  voltage  readings  are  also  given  the  deter- 
mination of  cos  0  is  simplified. 

True  power  Wi  ±  W\ 

Power  factor  =  cos  *  =  Apparent-power  - 


The  following  readings  taken  on  a  three-phase  induction  motor 
running  light  show  the  value  of  the  watt  ratio  curve. 


E 

I 

Wi 

W2 

W\ 
WI 

Cos  <j>  = 
Wz  ±Wi 

Cos  <f>  from 
Fig.  138 

Correct 
value 
of  TFi 

VZEI 

20 

1.01 

+4 

19 

+0.21 

0.656             0.66 

+4 

-4 

-0.21 

0.428 

0.35 

40 

1.24 

+5 

40 

+0.125 

0.528 

0.60 

o 

-0.125 

0.410 

0.41 

-5 

In  the  first  case  the  wattmeter  readings  must  be  added  while  in 
the  second  they  must  be  subtracted. 

It  is  possible  to  determine  the  total  power  in  a  balanced 
three-phase  circuit  with  one  wattmeter  without  opening  the 
circuit  to  change  the  current  coil  from  one  line  to  the  other. 
This  may  be  done  in  two  ways.  (1)  Referring  to  Fig.  136(6), 
first  connect  the  current  coil  in  line  A  with  the  potential  coil 
between  A  and  B,  the  reading  is 


=  EABIA  cos  (30  + 


V3EI  cos  (30  +  0); 


then  with  the  current  coil  still  in  line  A  connect  the  potential 
coil  from  B  to  C;  the  reading  is 


W 


EACIA  cos  (30  - 


and  the  power  is  the  sum  of  the  tw«   readings. 


POLYPHASE  ALTERNATING-CURRENT  CIRCUITS    163 


2.  The  second  method  requires  an  ammeter  and  voltmeter 
in  addition  to  the  wattmeter.  Connect  the  wattmeter  with  its 
current  coil  in  line  A  and  its  potential  coil  from  B  to  C;  the 
reading  is 

W  =  EBCIA  cos  (90  -  0),  Figs.    136(c)    and    137, 

=  EJi  sin  <f>. 


Sin 


W 


is  therefore  known  and  cos  <    can  be  calculated. 


The  power  in  the  circuit  is  ^3EtIi  cos  </>. 

This  latter  method  does  away  with  the  difficulty  encountered 
in  the  case  of  low  power  factors. 

The  two  wattmeters  connected  as  in  Figs.  136 (a)  and  139, 
give  the  total  power  in  the  circuit,  whether  the  loads  on  the 


B  i, 


FIG.  139. 


phases  are  balanced  or  unbalanced.  Take  the  case  of  the  delta- 
connected  circuit,  Fig.  139,  and  consider  the  instantaneous  values 
of  voltages  and  currents  and  power. 


e2A  = 


e22  -  e2i 


and 
but 


wz  =  e2i2  +  e3^3  — 
+  ez  +  63  =  0,  and 


—  (ez  +  e3), 


therefore   w\  +  ^2  =  v\i>\  +  e2i2  +  esis  =  the    sum    of    the   in- 
stantaneous values  of  power  in  the  three  phases. 

When  measuring  power  in  a  three-phase,  four-wire  system,  the 
two-wattmeter  method  will  give  accurate  results  only  when  the 
loads  are  balanced.  For  unbalanced  loads  three  wattmeters 
must  be  used  connected  in  the  three  lines  and  with  their  potential 
coils  connected  between  the  lines  and  neutral. 


164  ELECTRICAL  ENGINEERING 

126.  Example. — A  220-volt,  three-phase  alternator  supplies  the  following 
loads : 

1.  Twenty-five  kilowatts  at  100  per  cent,  power  factor  for  lighting. 

2.  A  100-hp.  (output)  induction  motor  of  92  per  cent,  efficiency  and  90 
per  cent,  power  factor. 

3.  A  number  of  small  induction  motors  with  a  combined  output  of  50  hp., 
average  efficiency  =  80  per  cent,  and  average  power  factor  =  75  per  cent. 

If  the  loads  on  the  three  phases  are  balanced,  find: 
(a)  The  output  of  the  alternator  in  kilowatts, 
(fe)  The  line  current, 
(c)   The  power  factor  of  the  combined  load. 

(a)  The  output  of  the  alternator  =  25.0  +  Q^  x  i  QQQ  +  Q~80  X  FOOO 

=  25.0  +  81.2  +  46.7  =  152.9  kw. 

25  X  1  000 

(6)  The  in-phase  current  per  line  for  (1)     =  —= — '— —  =  66  amp. 

\/o  X  2i2i(j 

The  quadrature  current  per  line  for  (1)  =0 

SI  2  X  1  000 

The  in-phase  current  per  line  for  (2)     =      '     — £— -  =  213  amp. 

\/o  X  £2\j 

213 

The  total  current  per  line  for  (2)  =  -pr^r  =  237  amp. 

u.y 

The  quadrature  current  per  line  for  (2)  =  V(237)2  -  (213)2  =  103  amp. 

46  7X1  000 

The  in-phase  current  per  line  for  (3)     =  — ^= j— —  =  123  amp. 

*\/3  X  220 

123 
The  total  current  per  line  for  (3)  =  Q-^.  =  164  amp. 

The  quadrature  current  per  line  for  (3)  =  V(164)2  -  (123)2  -  108  amp. 
The  total  in-phase  current  per  line  =  66  +  213  +  123  =  402  amp. 
The  total  quadrature  current  per  line  =  0  +  103  +  108  =211  amp. 
The  current  per  line  =  V(402)2+  (211)2  =  455  amp. 

402 
(c)  The  power  factor  of  the  load  is         X  100  =  88.3  per  cent. 


CHAPTER  VII 
DIRECT-CURRENT  MACHINERY 

127.  The  Direct-current  Dynamo. — A  direct-current  dynamo 
consists  of  an  electric  circuit,  connected  to  a  commutator  and 
tapped  by  brushes,  revolving  in  a  magnetic  field  which  is  produced 
by  stationary  electric  circuits. 

Such  a  machine  is  illustrated  in  Fig.  140  and  comprises  the 
following  parts: 

1.  Yoke  1 

2.  Pole  pieces  >  Magnetic  circuit. 

3.  Armature  core    J 

4.  Armature  winding    )  ^.       ,   .        ,     ,  .  ., 
_    ~                                >  Revolving  electric  circuit. 

5.  Commutator  J 

6.  Brushes  and  brush  holders.     Collecting  apparatus. 

7.  Field  winding.     Stationary  exciting  circuit. 

128.  Yoke. — The  yoke  serves  mechanically  as  the  frame  of  the 
machine  and  magnetically  to  carry  the  flux  from  pole  to  pole;  in 
small  machines  it  is  usually  made  of  cast  iron  but  in  machines 
where  great  weight  is  objectionable  it  is  made  of  cast  steel  which 
has  greater  strength  and  permeability. 

129.  Pole  Pieces. — The  pole  pieces  or  pole  cores  are  usually 
made  of  cast  steel  or  sheet  steel  and  are  bolted  to  the  yoke.     For 
small  machines  the  yoke  and  poles  are  sometimes  cast  in  one 
piece.     All  solid  poles  must  have  laminated  pole  faces  bolted  to 
them  in  order  to  reduce  the  eddy-current  loss  due  to  local  varia- 
tions of  the  magnetic  density  in  the  pole  faces  as  the  armature 
teeth  move  across  them. 

The  pole  cores  carry  the  field  windings  of  the  machine.  Solid 
poles  are  made  circular  and  so  have  the  greatest  section  for  a 
given  perimeter  and  require  the  smallest  length  of  field  copper. 
Laminated  poles  must,  however,  be  made  rectangular. 

The  section  of  the  pole  face  is  made  much  greater  than  that  of 
the  pole  core  in  order  to  reduce  the  flux  density  in  the  air  gap. 

165 


166 


ELECTRICAL  ENGINEERING 


DIRECT-CURRENT  MACHINERY 


167 


130.  Armature  Core. — The  armature  core  carries  the  rotating 
electric  circuit  in  slots  punched  out  on  its  periphery.     It  is  built 
up  of  sheets  of  steel  about  0.014  in.  in  thickness.     Alternate 
sheets  are  coated  with  an  insulating  varnish  to  increase  the  re- 
sistance in  the  path  of  the  induced  eddy  currents.     Open  spaces 
are  left  in  the  core,  called  vent  ducts  (V.V.,  Fig.  140),  which 
allow  air  to  circulate  through  the  armature  and  carry  off  the  heat 
generated  due  to  the  iron  and  copper  losses.     The  number  of 
vent  ducts  required  depends  on  the  length  of  the  armature. 

The  armature  punchings  are  carried  on  a  spider  S  and  are  kept 
in  place  by  heavy  end  plates  which  have  projections  on  their  outer 
edges  to  support  the  end  connections  of  the  armature  coils. 

131.  Armature  Winding. — The  armature  winding  is  the  seat 
of  the  generated  e.m.f.     It  must  be  tapped  at  certain  points  by 
brushes,  in  order  that  the  machine  may  supply  power  to  an 
external  receiver  circuit.     The  winding  consists  of  a  number  of 
coils  of  one  or  more  turns,  connected  together  to  form  a  con- 
tinuous winding;  leads  are  run  from  their  junctions  to  the  com- 
mutator bars  from  which  the  current  is  collected  by  the  brushes. 
The  coils  forming  the  winding  must  be  so  connected  together 
that  the  e.m.fs.  generated  in  coils  between  brushes  of  opposite 
polarity  will  all  act  in  the  same  direction. 


FIG.  141. — Bipolar  ring  winding. 

The  earliest  type  of  armature  winding  was  the  ring  winding, 
Figs.  141  and  142,  but  this  has  been  replaced  by  the  various  forms 
of  drum  windings,  a  few  of  which  are  illustrated  in  Figs.  144  to  151. 

132.  Ring  Windings. — In  the  bipolar  ring  winding,  Fig.  141, 
all  the  conductors  on  each  half  of  the  armature  are  connected  in 
series  between  the  brushes.  When  the  brushes  are  placed  on  the 
neutral  line,  that  is,  in  such  a  position  that  the  coil  being  corn- 
mutated  is  not  generating  any  e.m.f.,  the  e.m.fs.  generated  in  all 
conductors  under  one  pole  will  act  in  the  same  direction  and  will 
combine  to  give  the  terminal  e.m.f.  of  the  generator.  The  e.m.fs. 
generated  under  the  other  pole  will  be  equal  in  magnitude  but 


168 


ELECTRICAL  ENGINEERING 


will  act  in  the  opposite  direction.  Thus,  there  is  no  e.m.f .  tending 
to  cause  current  to  circulate  through  the  winding  at  no  load  and 
there  are  two  paths  in  multiple  for  the  current  flowing  through 
the  armature. 

The  connection. from  one  conductor  to  the  next  is  run  through 
inside  the  armature,  where  it  cannot  cut  magnetic  flux,  and  con- 
sequently one-half  of  the  winding  is  not  effective  in  generating 
e.m.f.  This  extra  wire  increases  the  resistance  of  the  armature 
and  adds  to  the  weight  and  cost  of  the  machine.  The  ring  wind- 
ing has  the  further  disadvantage,  that  it  is  very  difficult  to  replace 
injured  coils.  On  the  other  hand,  the  voltage  between  adjacent 
coils  is  so  low  that  very  little  insulation  is  required  between  them. 
This  type  of  winding  is  now  obsolete. 


FIG.  142. — Six-pole  ring  winding. 

Fig.  142  shows  a  six-pole  ring  winding  with  36  coils  connected 
to  a  commutator  with  36  bars.  This  winding  must  be  tapped  at 
six  equidistant  points  by  brushes;  there  are  six  paths  in  parallel 
through  the  armature  from  positive  to  negative  terminals  and  the 
voltage  of  the  machine  is  that  generated  in  one-sixth  of  the 
winding. 


FIG.   143.- 


O) 

-Armature  coils. 


133.  Drum  Winding. — In  drum-wound  machines  the  whole  of 
the  armature  winding  is  carried  in  slots  on  the  outside  of  the 
armature  core  and  both  sides  of  any  coil  are  effective  in  generat- 


DIRECT-CURRENT  MACHINERY 


169 


ing  e.m.f.     The  single  coils  are  of  the  shapes  shown  in  Fig.  143 
and  may  consist  of  one  or  more  turns. 

The  conductors  forming  the  two  sides  of  a  coil  must  be  situated 
in  fields  of  opposite  polarity  in  order  that  the  e.m.fs.  generated 

Generator 
I   -^ ^^  1 

U-^-J 


Equalizer  Rings 


I        ,  ,.,         t                .-          1  — 

N 

S 

N 

S 

N 

S 

FIG.  144  AND  145. — Six-pole  multiple-drum  winding. 

in  them  may  act  in  the  same  direction.  One  side  of  a  coil  is 
placed  in  the  top  of  a  slot  and  the  other  side  in  the  bottom  of  a 
slot  in  a  similar  position  under  the  next  pole. 


170 


ELECTRICAL  ENGINEERING 


According  to  the  way  in  which  the  end  connections  are  brought 
out  to  the  commutator  bars  and  the  coils  are  connected  together, 
drum  windings  are  divided  into  two  classes,  multiple  or  lap  wind- 


Generator 


- 


LJ- 


FIG.  146. — Six-pole  multiple-drum  winding.     Fractional  pitch. 

ings,  as  illustrated  by  coils  a  and  b  in  Fig,  143  and  the  windings  in 
Figs.  144  to  146  and  series  or  two-circuit  windings,  as  illustrated 
by  the  coils  c  and  d  in  Fig.  143  and  the  windings  in  Figs.  149  and 
150. 


DIRECT-CURRENT  MACHINERY  171 

134.  Multiple -drum  Windings. — In  the  multiple  winding  the 
two  terminals  of  a  coil  are  connected  to  adjacent  commutator 
bars.     Fig.   144  represents  a  multiple  winding  for  a  six-pole 
machine  with  72  conductors  and  36  slots.     The  sides  of  a  coil  are 
placed  in  slots  1  and  7  and  the  terminals  are  connected  to  bars 
1  and  2.     The  same  winding  is  shown  in  Fig.  145  and  the  direc- 
tions of  the  currents  are  shown  by  arrowheads.     The  brushes  are 
placed  on  the  no-load  neutral  points  and  therefore  directly  under 
the  centers  of  the  poles  and  as  many  sets  of  brushes  are  required 
as  there  are  poles. 

Tracing  through  the  winding  from  a  positive  to  a  negative 
brush  only  one-sixth  of  the  conductors  are  taken  and  there  are 
therefore  six  paths  in  multiple  through  the  armature  winding 
from  the  positive  to  the  negative  terminal  of  the  machine  and 
each  conductor  carries  only  one-sixth  of  the  current  passing 
through  the  armature.  The  number  of  paths  is  equal  to  the 
number  of  poles  as  in  the  ring  winding. 

Generally  in  multiple  windings  the  coil  pitch  is  almost  equal  to 
the  pole  pitch  and  the  windings  are  called  full-pitch  windings. 
When  the  coil  pitch  is  less  by  one  or  more  teeth  than  the  pole  pitch 
the  winding  is  called  a  fractional  pitch  or  short-chord  winding. 
Fig.  146  shows  a  fractional  pitch,  multiple-drum  winding.  Frac- 
tional-pitch windings  have  shorter  end  connections  than  full-pitch 
windings  and  have  a  smaller  inductive  e.m.f.  generated  during 
commutation  since  the  two  coils  in  one  slot  are  not  commutated 
at  the  same  time  and  thus  the  inductive  flux  is  that  due  to  half 
the  ampere-turns  acting  in  the  case  of  a  full-pitch  winding. 

135.  Equalizer  Rings. — In  multiple-wound  machines,  if  there 
is  any  irregularity  in  spacing  the  brushes  or  if  the  air  gaps  under 
all  the  poles  are  not  of  the  same  depth,  the  e.m.fs.  generated  in 
the  different  sections  of  the  winding  will  not  be  equal  and  the  un- 
balanced e.m.f.  will  tend  to  cause  current  to  circulate  through  the 
brushes  even  when  the  machine  is  not  carrying  any  load.     To 
keep  these  circulating  currents  out  of  the  brushes  similar  points 
under  the  different  pairs  of  poles  which  should  normally  be  at  the 
same  potential  are  joined  together  by  heavy  copper  connections 
called  equalizer  rings  and  these  provide  a  path  of  very  low  re- 
sistance which  the  circulating  currents  follow  in  preference  to  the 
comparatively  high-resistance  path  through  the  brushes  (Figs. 
144  and  145). 

Equalizer  rings  should  be  provided  on  all  multiple-wound 


172 


ELECTRICAL  ENGINEERING 


machines  except  the  smallest  sizes.  They  are  connected  to  every 
third  or  fourth  coil.  Windings  to  be  used  with  equalizer  rings 
must  have  the  number  of  coils  a  multiple  of  the  number  of  pairs 
of  poles. 

Fig.  147  shows  the  circulating  currents  in  a  four-pole  machine, 
in  which  the  gaps  under  the  two  lower  poles  have  been  shortened, 
due  to  wear  of  the  bearings.  The  voltages  generated  in  the  two 
lower  sections  of  the  winding  are  assumed  to  be  10  per  cent, 
higher  than  those  in  the  upper  sections.  Fig.  148  shows  three 
equalizer  rings  connected  to  this  machine  to  keep  the  circulating 
currents  out  of  the  brushes. 


FIG.  148. 

136.  Series-drum  Windings. — In  the  series  winding  the  ter- 
minals of  a  coil  are  connected  to  two  commutator  bars  approxi- 
mately twice  the  pole  pitch  apart.  Fig.  149  represents  a  series  or 
two-circuit  winding  for  a  six-pole  machine  with  44  conductors 
and  22  slots.  One  side  of  a  coil  is  placed  in  the  top  of  slot  1  and 
the  other  side  in  the  bottom  of  slot  5  and  the  terminals  of  the 
coil  are  connected  to  commutator  bars  1  and  8. 

Tracing  out  the  winding  from  the  positive  brush  BI  to  the  nega- 
tive brush  Bz  one-half  of  the  armature  conductors  are  taken  in. 
There  are  therefore  but  two  circuits  in  multiple  between  terminals 
independent  of  the  number  of  poles  and  the  winding  is  called  a 
two-circuit  or  series  winding. 

Only  two  sets  of  brushes  are  required  to  collect  the  current  but 
when  the  current  is  large  it  is  usual  to  employ  other  sets  of 
brushes  as  shown  at  Bz,  #4,  B$  and  J56  and  as  many  sets  of  brushes 
as  there  are  poles  may  be  used. 

Series  windings  are  used  in  small  high-voltage  machines  or 
where  it  is  desirable  to  use  only  two  sets  of  brushes,  as  in  small 
railway  motors;  but  in  large  multipolar  machines  with  many  sets 


DIRECT-CURRENT  MACHINERY 


173 


of  brushes  the  current  does  not  divide  equally  between  brushes 
of  the  same  polarity  and  commutation  is  unsatisfactory. 

The  number  of  coils  in  a  series  winding  must  be  one  more  or  one 
less  than  a  multiple  of  the  number  of  pairs  of  poles,  or 

N  =  C\  ±  1, 


LJT_     g-       g+      y-       g+       y- 

f                        r 

N 

s 

N 

S 

N 

S 

FIG.   149. — Six-pole  series-drum  winding,  retrogressive. 


where 


N  =  number  of  armature  coils, 
p/2  =  numbers  of  pairs  of  poles, 

and       C  =  a  constant  whole  number  =  avg.  winding  pitch. 
Each  coil  may  have  any  number  of  turns. 


174 
If 


ELECTRICAL  ENGINEERING 


AT=C|-1,  (241) 

the  winding  starting  from  bar  1  goes  once  around  the  armature 
and  is  connected  to  bar  2.  It  is  therefore  called  a  progressive 
winding  (Fig.  150). 


Generator 


FIG.  150. — Six-pole  series-drum  winding,  progressive. 

If  w-cjj+i, 


(242) 


the  winding  starting  from  bar  1  and  going  once  around  the  arma- 
ture is  connected  to  the  bar  before  1  and  it  is  called  a  retrogres- 
sive winding  (Fig.  149). 

137.  Double  Windings. — If  space  is  left  between  adjacent 
coils  of  a  multiple  winding,  a  second  winding  may  be  placed  on 


DIRECT-CURRENT  MACHINERY 


175 


the  same  core.  The  second  winding  may  be  entirely  separate 
from  the  first,  that  is,  each  of  the  two  windings  closed  upon  itself; 
or  after  passing  through  the  first  winding  the  circuit  enters  the 
second  and  after  passing  through  the  second  reenters  the  first. 


Generator 


r  —  i 


FIG.  151. — Sixpole  multiple-drum  winding,  duplex,  doubly  reentrant. 

In  the  first  case  the  winding  is  duplex  doubly  reentrant  and  in  the 
second  case  duplex  singly  reentrant.  Duplex  multiple  windings 
have  twice  as  many  circuits  in  multiple  between  terminals  as  there 
are  poles.  Such  windings  are  suitable  for  large  low-voltage  ma- 


176 


ELECTRICAL  ENGINEERING 


chines  used  in  electrolytic  work.  The  brushes  must  be  wide 
enough  to  collect  current  from  both  sections  of  the  winding  at  the 
same  time.  Fig.  151  shows  a  duplex  doubly  reentrant  winding 
for  a  six-pole  machine  with  72  conductors  and  36  slots. 

Similarly  the  series  winding  may  be  made  double  by  placing  a 
second  winding  in  alternate  slots  and  connecting  it  to  alternate 
commutator  bars.  The  second  winding  is  in  multiple  with  the 
first  and  there  are  four  paths  in  multiple  between  terminals. 

138.  Commutator. — The  commutator  is  one  of  the  most  im- 
portant parts  of  a  direct-current  machine.  It  consists  of  a  num- 
ber of  bars  of  hard-drawn  copper,  insulated  from  one  another  by 
thin  sheets  of  mica  or  other  insulating  material,  and  built  up  into 
the  form  of  a  cylinder  (Fig.  152).  The  bars  are  held  together  by 


Micanite 


FIG.   152. — Commutator. 


a  cast-iron  spider  from  which  they  are  insulated  by  micanite  "  V" 
rings.  The  terminals  of  the  coils  forming  the  armature  winding 
are  connected  to  the  bars  either  directly  by  soldering  them  into 
slots  in  the  bars  or  by  means  of  vertical  connectors  called  commu- 
tator risers.  In  order  that  the  brushes,  which  collect  the  current 
from  the  commutator,  may  run  smoothly  without  vibrating  or 
chattering  the  commutator  surface  must  be  perfectly  round  and 
smooth. 

The  function  of  the  commutator  is  illustrated  in  Fig.  153.     The 
current  from  the  machine  is  I  amp.  and  the  current  in  each 

conductor  is  Ic  =  ^  amP-  During  the  time  taken  for  the  brush 
to  move  across  the  insulation  between  bars  1  and  2  the  current 


DIRECT-CURRENT  MACHINERY 


177 


in  coil  c  must  change  from  Ic  in  one  direction  to  Ic  in  the  opposite 
direction.  This  reversal  of  the  current  is  called  commutation  and 
to  be  satisfactory  it  must  be  effected  without  sparking.  The 
brushes  are  shown  placed  on  the  neutral  line  and  the  coils  short- 
circuited  are  not  cutting  any  flux  and  therefore  have  no  e.m.f. 
generated  in  them  due  to  rotation.  If  the  short-circuited  coil  had 
no  inductance  the  current  would  reverse  completely  due  to  the 
contact  resistance  between  the  brush  and  the  commutator. 


deb 


i  i          i  3  i  2 


U 

2lc 

Q 


FIG.   153. — Commutation. 


In  (1)  the  current  in  coil  c  is  7C;  in  (2)  it  is  zero  since  current  Ic 
from  bar  1  goes  through  one-half  of  the  brush-contact  area  and 
current  Ic  from  bar  2  goes  through  the  other  half,  and  the  drop  of 
voltage  on  both  sides  is  the  same  and  therefore  there  is  no  voltage 
available  to  drive  the  current  through  the  resistance  of  the  coil. 
Between  (1)  and  (2)  the  resistance  from  bar  2  to  the  brush  is 
greater  than  the  resistance  from  bar  1  to  the  brush  and  so  part  of 
the  current  from  2  flows  through  the  coil  c.  Between  (2)  and  (3) 
the  resistance  from  bar  2  to  the  brush  is  less  than  from  bar  1  and 
part  of  the  current  from  1  flows  through  the  coil  c.  In  (3)  the 
current  in  c  is  Ic  but  in  the  opposite  direction  from  that  in  (1) 
and  commutation  is  complete. 

The  self -inductance  L  of  the  coil  opposes  any  change  of  current 

by  generating  a  back  e.m.f.  L  -r  volts;  when  the  current  is  large 

and  the  time  of  commutation  short  this  back  e.m.f.  is  large  and 
the  current  will  not  be  reversed  when  the  brush  breaks  contact 
with  bar  1  and  sparking  will  occur. 

To  counteract  the  effect  of  self-inductance  the  brushes  in  a  gen- 
erator are  moved  ahead  of  the  neutral  in  the  direction  of  rotation 
and  back  in  a  motor.  The  short-circuited  coil  is  then  in  a  field 
which  generates  in  it  an  e.m.f.  due  to  rotation  which  opposes  the 

back  e.m.f.  of  self-inductance,  or,  as  it  is  usually  called,  "the  re- 
12 


178 


ELECTRICAL  ENGINEERING 


actance  voltage  of  the  coil,"  and  assists  commutation.  The  prob- 
lem of  commutation  is  discussed  fully  in  Art.  192. 

139.  Brushes  and  Brush  Holders. — The  brushes  collect  the 
current  from  the  moving  commutator  and  from  them  it  passes  to 
the  receiver  circuit. 

Brushes  were  at  first  made  of  copper  because  it  had  a  low  re- 
sistance and  large  current-carrying  capacity  but  commutation  of 
large  currents  was  not  satisfactory.  Carbon  brushes  were  then 
introduced  and  commutation  was  greatly  improved  due  to  the 
action  of  the  high-resistance  contact  film  between  the  brush  and 
commutator.  A  much  better  contact  surface  was  also  obtained 
and  the  wear  on  the  commutator  was  reduced.  But  since  carbon 
will  carry  only  about  30  to  50  amp.  per  square  inch  while  copper 
will  carry  150  to  200  amp.  per  square  inch  a  much  larger  brush 
area  is  required  and  a  larger  commutator. 

In  order  to  maintain  a  good  contact  between  the  brush  and 
commutator  a  spring  is  used  exerting  on  the  brush  a  pressure  of 
about  1^2  "to  2  Ib.  per  square  inch  of  contact  area. 

The  brush  holders  are  made  of  brass  and  carry  part  of  the  cur- 
rent but  leads  are  connected  directly  from  the  brushes  to  the  main 
leads  of  the  machine  to  prevent  any  drop  of  voltage  which  might 
occur  due  to  poor  contact  between  brushes  and  holders. ' 


Load 


Source  of  „  .        . 

Power  1  (      I      Load 

* 

FIG.  154. — Separate  excitation. 


Load 


0 


Load 


FIG.   155. — Shunt  or  self-excitation. 


140.  Field  Windings. — The  field  winding  is  a  stationary  .elec- 
tric circuit  consisting  of  one  or  more  coils  of  wire  placed  on 
each  of  the  field  poles.  They  are  supplied  with  current  and  pro- 
vide the  m.m.f.  necessary  to  drive  the  magnetic  flux  through  the 


DIRECT-CURRENT  MACHINERY 


179 


machine.  The  methods  used  in  calculating  the  number  of 
ampere-turns  required  to  produce  the  flux  in  a  machine  is  worked 
out  in  Art.  227. 

According  to  the  manner  of  exciting  the  fields,  direct-current 
machines  are  divided  into  magneto  machines  in  which  the 
flux  is  produced  by  permanent  magnets;  separately  excited  ma- 
chines (Fig.  154)  in  which  the  flux  is  produced  by  a  winding  sup- 
plied with  current  from  some  source  outside  the  machine;  shunt 
machines  in  which  the  field  winding  is  connected  across  the  arma- 
ture terminals  and  receives  a  small  current  at  the  full-machine 
voltage  (Fig.  155) ;  series  machines  in  which  the  field  winding  is 
connected  in  series  with  the  armature  and  carries  the  full  arma- 
ture current  (Fig.  156);  and  compound  machines  in  which  the 
field  has  both  a  shunt  and  a  series  winding  (Fig.  157). 


FIG.  156. — Series  excitation. 


<a> 

(a)  Long  Shunt  (6)   Short  Shunt 

FIG.   157. — Compound  excitation. 


According  to  the  number  of  poles  machines  are  divided  into 
bipolar  and  multipolar  machines  but  the  bipolar  type  is  adapted 
only  for  small  sizes. 

141.  Direction  of  Rotation  of  Generators  and  Motors. — Fig. 
158  represents  either  a  generator  or  motor.  The  directions  of  the 
currents  in  the  armature  are  shown  by  the  dots  and  crosses  and 
the  directions  of  rotation  by  arrows. 

With  currents  as  shown,  the  direction  of  rotation  of  a  generator 
is  counter-clockwise  and  of  a  motor  is  clockwise. 

These  results  are  obtained  by  examining  the  fields  produced  by 
the  armature  currents. 


180  ELECTRICAL  ENGINEERING 

Take  for  example  the  conductor  A.  Its  field  combines  with 
the  main  field  and  produces  a  strong  field  below  the  conductor 
and  a  weak  field  above  it.  There  is  therefore  a  force  /  acting  on 
the  conductor  tending  to  move  it  up.  This  is  the  force  that  must 
be  overcome  by  the  engine  driving  the  generator  in  order  to  de- 
velop electric  power  and  therefore  the  rotation  of  a  generator  is 
against  this  force  and  is  counter-clockwise. 

In  the  case  of  the  motor,  the  force  /  on  the  conductor  is  the  me- 
chanical force  developed  and  the  rotation  is  in  the  direction  of  this 
force  and  is  clockwise. 

To  reverse  the  direction  of  rotation  of  a  motor  it  is  necessary  to 
reverse  either  the  amature  current  or  the  field  current  but  not 
both. 


Motor 
FIG.  158.  —  Direction  of  rotation. 

142.  Generation  of  Electromotive  Force.  —  The  e.m.f  .  generated 
in  the  armature  of  a  direct-current  generator  or  motor  is 

8  =  Zn3>  2-  10-8  volts, 
Pi 

where  Z  =  number  of  conductors  on  the  armature, 
n    —  speed  of  armature  in  r.p.s., 
$    =  flux  crossing  the  air  gap  from  one  pole, 
p    =  number  of  poles, 

and      pi  =  number  of  paths  in  multiple  between  terminals. 

In  1  sec.  each  conductor  cuts  n$p  lines  of  force  and  thus  the 
average  e.m.f.  generated  in  each  of  the  Z  conductors  is 

e  =  n®    10~8  volts. 


Between  the  terminals  there  are  —  conductors  connected  in  series 

Pi 
and  therefore  the  e.m.f.  between  terminals  is 

8  =  Zn$>  2-  10-8  volts.  (243) 

Pi 


DIRECT-CURRENT  MACHINERY 


181 


This  is,  the  e.m.f.  equation  of  a  direct-current  generator;  it  may 
be  written 

(244) 


where 


Q        

np 

K  =  Z  —  10~8  is  a  constant  of  the  machine. 
Pi 


The  e.m.f.  is  therefore  directly  proportional  to  the  speed  and  to 
the  flux  crossing  the  air  gap. 

Fig.  159  shows  the  relation  between 
e.m.f.  and  speed,  for  a  constant  value 
of  <l>.  The  locus  is  a  straight  line 
passing  through  the  origin. 

In  a  generator  the  current  flows  in 
the  direction  of  the  generated  e.m.f. 
but  in  a  motor  an  e.m.f.  is  impressed 
on  the  terminals  of  the  armature  and 
drives  a  current  against  the  e.m.f. 
generated  in  the  motor.  The  e.m.f. 


100 

^9  76 
h 

3 
^50 

s» 

O 
0 

z 

/ 

'/ 

Z 

1            25            50           75         100 
Speed  n 

FIG.  159. — Variation  of  gener- 
ated e.m.f.  with  speed. 


generated  in  a  motor  armature  is  therefore  called  a  back  e.m.f. 
143.  Effect  of  Moving  the  Brushes.  —  The  equation 


8  = 


-  10 
Pi 


Maximum  Voltage 
Generator 


Generator 


holds  only  if  the  brushes  are  on  the  no-load  neutral  points. 
When  the  brushes  are  moved  ahead  of  the  neutral  points  or 
behind  them  the  e.m.f.  between  terminals  is  decreased.  This 
may  be  seen  by  reference  to  Fig.  160.  With  the  brushes  on 

the  neutral  points  the 
e.m.fs.  generated  in  all 
the  conductors  in  series 
between  terminals  act  in 
the  same  direction  and 
combine  to  give  the 
maximum  e.m.f.  Wlien 
the  brushes  are  moved 

i_          i        j_v  r         v 

ahead     the     e.m.f.     be- 

FIG.  160.—  Effect  of  moving  the  brushes.        tween  terminals  is  only 

that   generated   in  con- 

ductors a-b  or  d-c  since  the  resultant  of  the  e.m.fs.  generated 
in  conductors  a-d  and  b-c  is  zero.  Thus  advancing  the  brushes 
corresponds  to  a  decrease  in  the  number  of  armature  conductors. 


182  ELECTRICAL  ENGINEERING 

144.  Building  up  of  Electromotive  Force  in  a  Self -excited 
Generator. — In  a  self-excited  generator  at  rest  there  is  no  flux 
crossing  the  gap  except  the  residual  magnetism.     When  the 
armature  is  rotated  only  a  small  e.m.f.  is  generated  in  it  and  a 
very  small  current  is  produced  in  the  field  winding.     If  the 
m.m.f .  of  this  current  is  in  the  direction  of  the  residual  magnetism, 
it  will  increase  the  flux  and  the  e.m.f.  will  increase  and  gradually 
build  up  to  its  full  value. 

If,  however,  the  current  opposes  the  residual  magnetism,  it  will 
cause  it  to  decrease  and  the  e.m.f.  will  not  build  up  until  the  field 
winding  is  reversed. 

If  there  is  no  residual  magnetism,  the  e.m.f.  cannot  build  up 
until  power  is  supplied  to  the  winding  from  some  outside  source  to 
start  the  flux. 

145.  Armature  Reaction  and  Distribution  of  Magnetic  Flux. — 
Fig.  161  shows  approximately  the  distribution  of  flux  in  a  two- 
pole  machine  with  the    fields  excited  but  without  current  in  the 


t!urve(2),   B 
Curve  (1),   My 


)OOOOOOOOOOOOOOOC>QgOOOOOOOOOOOOOOOOji 

-i  ^^  -      \r. 


(a)  (6) 

FIQ.  161. — Distribution  of  flux  and  m.m.f.  at  no  load. 

armature  winding.  Curve  1  shows  the  m.m.f.  acting  at  each 
point  of  the  armature  circumference;  under  the  north  pole  it  is 
positive  and  has  a  constant  value;  under  the  south  pole  it  is 
negative  but  of  the  same  magnitude;  beyond  the  pole  tips  its 
value  may  be  represented  by  the  straight  line  which  passes 
through  zero  midway  between  the  poles.  The  m.m.f.  is  expressed 
in  ampere-turns  and  is  denoted  by  M/. 

The  flux  density  produced  by  this  m.m.f.  is  at  every  point  di- 
rectly proportional  to  the  m.m.f.  and  is  inversely  proportional  to 
the  reluctance  of  the  path.  It  is  of  constant  value  over  the  pole 
face  if  the  air  gap  is  uniform  but  falls  off  rapidly  beyond  the  pole 
tips  due  to  the  increased  reluctance  of  the  air  path  and  to  the 
decrease  in  the  m.m.f.  acting.  Midway  between  the  poles  it  is 
zero.  It  is  represented  by  B  and  its  values  are  plotted  in  curve 
2.  The  total  flux  entering  the  armature  is  represented  by  the 


DIRECT-CURRENT  MACHINERY 


183 


area  under  curve  2  and  this  is  the  value  of  $  which  appears  in  the 
e.m.f,  equation. 

When,  however,  the  armature  is  carrying  current  it  exerts  a 
m.m.f.  called  armature  reaction,  which  combines  with  the 
m.m.f.  of  the  field  winding  and  changes  both  the  distribution 
and  the  total  value  of  the  flux  entering  the  armature. 


Curve  (1),M0 
Curve (2),  B 


FIG.   162. — Armature  m.m.f.  and  flux. 

Fig.  162  (a)  shows  the  distribution  of  flux  produced  by  the 
armature  m.m.f.  acting  alone,  and  the  values  of  the  m.m.f.  of  the 
armature  at  all  points  around  the  circumference  are  plotted  in 
curve  1,  Fig.  162(6).  The  brushes  are  placed  on  the  no-load 
neutral  points.  The  armature  m.m.f.  Ma  is  a  maximum  in  line 
with  the  brushes  and  falls  off  as  a  linear  function  to  zero  under 
the  center  of  the  poles.  The  distribution  of  the  flux  produced  by 
the  armature  m.m.f.  is  shown  in  curve  2,  Fig.  162(6). 

Curve(l), M-M/  +  M  a 
Curve  (2)    B 


FIG.  163. — Distribution  of  flux  and  m.m.f.  under  load,  with  the  brushes  on  the 
no-load  neutral  points. 

Fig.  163 (a)  represents  the  conditions  when  the  m.m.f s.  of  field 
and  armature  are  acting  together  and  with  the  brushes  still  on  the 
no-load  neutral  points.  Curve  1,  Fig.  163(6),  shows  the  resultant 
m.m.f.  acting  at  each  point.  Its  ordinates  are  represented  by  M 
and  they  are  the  sum  of  the  corresponding  ordinates  M/  and  M a. 

The  m.m.f.  across  the  pole  face  is  no  longer  constant  but  is 


184 


ELECTRICAL  ENGINEERING 


decreased  over  one-half  and  increased  over  the  other  half  by  the 
same  amount.  The  flux  density  B  (curve  2)  at  each  point  is  still 
proportional  to  the  m.m.f.  and  inversely  proportional  to  the 
reluctance  of  the  path,  but,  since  part  of  the  path  is  made  up  of 
a  magnetic  material,  due  to  the  effect  of  saturation,  the  increase 
of  flux  under  one-half  of  the  pole  is  less  than  the  decrease  under 
the  other  half  and  consequently  the  total  flux  is  decreased. 

If  it  were  not  for  the  effect  of  saturation  the  ordinates  of  curve 
2,  Fig.  163(6),  could  be  obtained  by  adding  the  corresponding 
ordinates  of  Fig.  161(6)  and  Fig.  162(6). 

The  neutral  points  are  no  longer  midway  between  the  poles  but 
have  been  shifted  in  the  counter-clockwise  direction  in  Fig.  163  (a) 
and  to  the  right  in  Fig.  163(6).  To  prevent  sparking  the  brushes 
must  be  moved  up  to  or  a  little  vbeyond  the  load  neutral  points. 
In  a  generator  the  brushes  must  be  moved  forward  in  the  direc- 
tion of  rotation  and  in  a  motor  must  be  moved  backward  against 
the  direction  of  rotation  as  indicated  in  Fig.  164. 

With  the  brushes  midway  between  the  poles  the  direction  of 
the  armature  m.m.f.  is  at  right  angles  to  the  field  m.m.f.  and  it 
therefore  does  not  directly  oppose  it  but  causes  a  distortion  of  the 
flux  and  a  decrease  due  to  saturation.  In  this  case  the  m.m.f.  of 
the  armature  is  cross-magnetizing. 


Generator 


Curved  ).M=M/+Ma 
Curve(2),  B 


FIG.   164. — Distribution    of    flux    and    m.m.f.  under  load,  with    the    brushes 
advanced  to  the  pole  tips. 

When,  however,  the  brushes  are  moved  into  the  fringe  of  lines 
at  the  pole  tips,  as  in  Fig.  164(a),  the  two  m.m.fs.  are  no  longer  at 
right  angles  and  as  seen  in  Fig.  164(6)  the  part  of  the  armature 
m.m.f.  which  is  subtracted  from  the  field  m.m.f.  is  much  greater 
than  the  part  which  is  added  to  it  and  therefore  the  resultant 
m.m.f.  is  reduced  and  the  flux  is  both  distorted  and  decreased. 

Referring  to  Fig.  164  (a)  the  armature  conductors  may  be 
separated  into  two  groups ;  namely  those  between  a  and  d  included 


DIRECT-CURRENT  MACHINERY  185 

in  the  double  angle  of  advance  a  with  their  return  conductors  from 
c  to  b  and  those  under  the  pole  between  b  and  a  with  their  return 
conductors  between  d  and  c.  Th,e  first  group  acts  in  direct 
opposition  to  the  field  m.m.f.  and  decreases  the  flux  crossing  the 
air  gap.  They  are  therefore  called  the  demagnetizing  ampere- 
turns  of  the  armature.  This  demagnetizing  m.m.f.  increases  as 
the  shift  of  the  brushes  is  increased  and  it  also  increases  directly 
with  the  armature  current. 

The  second  group  exerts  a  m.m.f.  at  right  angles  to  the  field 
m.m.f.  and  distorts  the  flux  as  in  Fig.  163 (a)  and  causes  a  de- 
crease due  to  saturation.  They  are  called  the  cross-magnetizing 
ampere-turns  of  the  armature. 

For  sparkless  commutation  without  the  use  of  interpoles  the 
brushes  must  be  moved  ahead  of  the  neutral  points  in  order  that 
the  coils  short-circuited  by  them  may  be  cutting  the  fringe  of  flux 
at  the  pole  tips.  E.m.fs.  are  thus  generated  in  the  coils  opposing 
the  back  e.m.fs.  due  to  inductance,  and  they  aid  in  reversing  the 
current.  As  the  armature  current  is  increased  a  point  is  finally 
reached  where  the  armature  m.m.f.  is  so  strong  that  it  over- 


FIG.   165. — Distribution  of  flux  and  m.m.f.  with  the  brushes  under  the  centers 

of  the  poles. 

balances  the  field  m.m.f.  at  the  pole  tips  and  therefore  no  revers- 
ing field  is  left  and  commutation  is  not  possible  without  interpoles. 
It  is  of  no  use  to  move  the  brushes  further  ahead  because  that  only 
increases  the  demagnetizing  component  of  armature  m.m.f.  and 
decreases  the  flux  more.  In  direct-current  machines  without 
interpoles  the  armature  ampere-turns  per  pole  at  full  load 
should  not  exceed  70  per  cent,  of  the  field  ampere-turns  per  pole. 
Fig.  165  shows  the  effect  of  moving  the  brushes  to  the  center 
of  the  poles.  The  whole  armature  m.m.f.  is  demagnetizing  and 
the  flux  is  reduced  to  a  small  value.  There  is  no  difference  of 
potential  between  the  brushes  since  the  sum  of  the  e.m.fs. 
generated  in  one-half  of  the  conductors  in  series  between  the 


186 


ELECTRICAL  ENGINEERING 


brushes  is  exactly  equal  and  opposite  to  that  generated  in  the 
other  half. 

146.  No-load  Saturation  Curve. — The  no-load  saturation 
curve  of  a  generator  shows  the  relation  between  the  voltage 
generated  at  no  load  and  the  field  current.  It  is  sometimes 
plotted  with  the  values  of  $g,  the  useful  flux  as  ordinates  and  the 
field  m.m.f.  or  field  ampere-turns  per  pole  as  abscissae. 

To  obtain  the  no-load  saturation  curve,  the  generator  is  run  at 
constant  speed  and  a  variable  resistance  is  connected  in  series 
with  the  field  winding.  As  the  current  is  raised  from  a  very 
low  value  the  flux  and  the  generated  voltage  increase  directly 
as  the  current  while  the  iron  parts  of  the  magnetic  circuit  are 
unsaturated,  but  as  the  flux  density  increases  the  magnetic  cir- 
cuit becomes  saturated  and  a  greater  increase  in  current  is  re- 
quired to  produce  a  given  increase  in  voltage  than  on  the  lower 
part  of  the  curve  (Fig.  166). 


o  a 

Field  Ampere  Turns 


o  0i  02  abed 

Field  Ampere  Turns 


FIG.  166. — No-load  saturation  curve.      FIG.  167. — Full-load  saturation  curve. 

Shunt-excited  machines  are  operated  at  a  point,  slightly  above 
the  knee  of  the  saturation  curve  to  insure  stability,  that  is,  in 
order  that  slight  changes  of  speed  may  not  cause  large  changes  in 
voltage. 

Machines  with  compound  windings  must  be  operated  at  a 
lower  point  on  the  saturation  curve  in  order  that  the  required  in- 
crease of  voltage  may  be  obtained  without  supplying  too  large 
a  m.m.f.  in  the  series  field  windings. 

If  the  field  current  is  gradually  decreased  from  its  maximum 
the  flux  and  the  generated  voltage  fall  off,  but  they  come  down  on 
a  curve  slightly  above  the  original  curve,  which  cuts  the  vertical 
axis  at  a  point  above  the  origin.  The  value  of  flux  represented 
by  this  intercept  is  the  residual  magnetism  in  the  machine. 


DIRECT-CURRENT  MACHINERY  187 

147.  Load  Saturation  Curves. — The  curve  showing  the  rela- 
tion between  the  terminal  voltage  and  the  field  current,  when  the 
generator  is  delivering  a  given  load  current  is  called  a  load  satura- 
tion curve  (Fig.  167).  The  load  saturation  curves  may  be  ob- 
tained from  the  no-load  saturation  curve  if  the  values  of  the 
demagnetizing  and  cross-magnetizing  armature  ampere-turns 
are  known  and  also  the  resistance  of  the  armature. 

If  the  brushes  are  moved  ahead  of  the  no-load  neutral  position 
by  an  angle  a  electrical  degrees  the  demagnetizing  ampere-turns 

per  pole  =  jorj  X  armature  ampere-turns  per  pole  (Fig.  164). 

If  a  =  18  degrees  or  10  per  cent,  of  the  pole  pitch,  the  demagnet- 
izing ampere-turns  per  pole  =  0.2  X  armature  ampere-turns 
per  pole. 

To  overcome  the  effect  of  the  demagnetizing  ampere-turns  an 
equal  number  of  ampere-turns  must  be  added  to  the  field  winding. 
This  number  should  be  increased  15  or  20  per  cent,  to  make  allow- 
ance for  the  leakage  of  flux  between  the  pole  and  armature  (Art. 
219). 

The  remaining  armature  ampere-turns   per  pole  are  _cross- 

180  -  2a 

magnetizing.     They    are      — ir^r —  X  armature    ampere-turns 

loU 

per  pole,  or  when  a  =  18  degrees,  the  cross-magnetizing  turns  = 
0.8  X  armature  ampere-turns  per  pole. 

The  cross-magnetizing  ampere-turns  produce  a  very  high 
saturation  in  the  teeth  under  one-half  of  the  pole  tips  and  so  in- 
crease the  reluctance  of  the  magnetic  circuit  as  a  whole  and  de- 
crease the  flux.  This  effect  is  very  difficult  to  determine  accu- 
rately but  in  normally  designed  machines  in  which  the  field 
ampere-turns  per  pole  =  approximately  150  per  cent,  of  the 
armature  ampere-turns  per  pole,  the  cross-magnetizing  ampere- 
turns  may  be  replaced  by  40  per  cent,  of  their  number  of  demag- 
netizing ampere-turns. 

This  value  can  be  applied  only  for  points  on  the  no-load  satura- 
tion curve  just  above  the  knee.  Where  the  saturation  is  very 
high  the  effect  of  the  cross-magnetizing  turns  is  relatively  larger 
and  a  larger  percentage  than  40  must  be  used.  On  the  other 
hand,  for  points  below  the  knee  of  the  saturation  curve  much 
smaller  percentages  must  be  used. 

In  the  case  of  the  generator  designed  in  Chapter  VIII,  the  arma- 
ture ampere-turns  per  pole  =  4,860  and  the  angle  of  brush  lead 


188  ELECTRICAL  ENGINEERING 

a  =  18  electrical  degrees;  the  demagnetizing  ampere-turns  per 
pole,  =  0.2  X  4,860  =  972,  and  the  cross-magnetizing  turns 
=  0.8  X  4,860  =  3,888.  Here  the  full  voltage  point  is  just  at 
the  knee  of  the  saturation  curve  and  30  per  cent,  of  3,888  =  1,166 
demagnetizing  ampere-turns  are  taken  to  represent  the  cross- 
magnetizing  effect. 

The  demagnetizing  effect  varies  directly  with  the  current  but 
the  cross-magnetizing  effect  increases  at  a  more  rapid  rate.  If 
the  two  effects  are  equal  at  full  load,  then  on  overloads  the  cross- 
magnetizing  effect  will  be  greater  than  the  demagnetizing  effect 
while  at  light  loads  it  will  be  less. 

In  Fig.  167,  os  is  the  no-load  saturation  curve  plotted  on  a  base 
of  field  ampere-turns.  At  no  load  the  field  m.m.f.  required  for 
rated  voltage  is  oa.  Let  ab  be  the  demagnetizing  ampere-turns  per 
pole  at  full  load  and  be  be  the  demagnetizing  ampere-turns  equiva- 
lent to  the  cross-magnetizing  turns.  In  order  to  generate  the  same 
voltage  under  load  as  at  no  load  the  field  m.m.f.  per  pole  must  be 
increased  by  the  amount  ac  =  mp  and  p  is  a  point  on  the  curve 
OiSi  showing  the  relation  between  the  voltage  generated  under 
load  and  the  field  m.m.f.  Curve  OiSi  would  be  parallel  to  os  if 
the  cross-magnetizing  effect  were  constant  but  at  the  lower  points 
on  the  curve  it  becomes  very  small  and  the  intercept  001  may  be 
taken  as  =  ab. 

The  terminal  voltage  E  will  be  less  than  the  generated  voltage 
8  by  the  resistance  drop  in  the  armature  and  series  field,  Ir  = 
pq.  The  curve  0282  is  drawn  parallel  to  the  curve  o^i  at  a  dis- 
tance pq  =  Ir  below  it.  This  is  the  full-load  saturation  curve  for 
the  generator. 

A  saturation  curve  for  half  load  could  be  obtained  by  joining 
the  center  points  of  all  the  lines  such  as  mq,  m^i,  etc. 

148.  Voltage  Characteristic  or  Regulation  Curve.  —  The  voltage 
characteristic  of  a  direct-current  generator  is  the  curve  showing 
the  relation  between  the  terminal  voltage  and  the  current  output 
or  load. 

The  voltage  generated  in  the  armature  is 


8  =  Zn<l>10-8  =  Kn$  volts.  (Art.  142) 

Pi 

Take  first  the  case  of  a  separately  excited  generator  in  which  the 
speed  n  and  the  field  current  //  are  both  kept  constant. 


DIRECT-CURRENT  MACHINERY  189 

At  no  load  the  flux  crossing  the  air  gap  under  each  pole  is  3>0 
and  the  voltage  generated  is 

80  =  Kn$Q] 

this  is  also  the  terminal  voltage  at  no  load. 

As  the  generator  is  loaded  the  terminal  vpltage  decreases  due 
to  two  causes:  (a)  armature  reaction,  and  (6)  armature  resist- 
ance. 

(a)  When  current  flows  in  the  armature,  the  armature  m.m.f., 
which  is  partly  demagnetizing  and  partly  cross-magnetizing, 
reduces  the  flux  crossing  the  gap  and  thus  reduces  the  generated 
voltage.  The  loss  of  flux  increases  at  a  more  rapid  rate  than  the 
current.  The  consequent  loss  of  voltage  is  called  the  drop  due 
to  armature  reaction. 

(6)  It  is  necessary  to  distinguish  between  the  voltage,  £, 
generated  in  the  armature  and  the  terminal  voltage,  E.  At  no 
load  they  are  the  same  but  when  current  flows  in  the  armature 
part  of  the  generated  voltage  is  consumed  in  driving  the  armature 
current  /  through  the  resistance  of  the  armature  winding  and 
brushes  r.  This  resistance  drop  is  Ir  and  increases  directly  with 
the  current. 

The  terminal  voltage  is          E  =  8  —  Ir.  (245) 

In  Fig.  168  are  shown  the  no-load  and  load  saturation  curves 
plotted  on  the  same  base.  The  constant  exciting  ampere-turns 
=  oa.  A  vertical  Kne  through  a  cuts  the  various  saturation 
curves  at  ra,  n,  r  and  t  which  represent  the  terminal  voltages  corre- 
sponding to  the  respective  loads.  These  points  are  plotted  on  a 
load  base  as  curve  m\r  and  this  is  the  voltage  characteristic  of 
the  generator.  It  is  sometimes  called  the  external  characteristic 
to  distinguish  it  from  the  corresponding  curve  of  generated 
voltages  niik. 

The  straight  line  ol  represents  the  armature  resistance  drops 
at  all  loads  =  Ir. 

The  voltage  generated  in  the  armature  at  any  load  is  8  =  E  + 
Ir  and  by  adding  the  ordinates  of  ol  to  those  of  m\r  the  curve  mik 
is  obtained  showing  the  relation  between  the  generated  voltage 
and  the  load.  This  is  called  the  internal  voltage  characteristic. 

At  full  load  the  total  drop  of  voltage  below  the  no-load  value 
is  mr  and  it  is  made  up  of  two  parts,  mk  the  drop  due  to  armature 
reaction  and  kr  the  armature  resistance  drop. 


190 


ELECTRICAL  ENGINEERING 


149.  Field  Characteristic  or  Compounding  Curve. — The  field 
characteristic  of  a  generator  is  the  curve  showing  the  increase  of 
field  current  required  under  load  to  maintain  a  constant  terminal 
voltage  at  constant  speed. 

In  a  separately  excited  or  self-excited  generator  the  terminal 
voltage  can  be  maintained  constant  as  the  load  current  increases 
if  the  field  current  is  increased  to  such  a  value  that  the  increase 
in  field  m.m.f.  will  not  only  overcome  the  effect  of  armature 
m.m.f.  but  will  produce  an  increase  in  flux  to  provide  the  extra 
voltage  to  supply  the  armature  resistance  drop. 


M  Load 


l^load 


Fia.  168. — Voltage  characteristics     FIG.  169. — Field  characteristics, 
separately  excited. 


Referring  to  Fig.  167  the  terminal  voltage  at  no  load  =  am 
and  the  field  m.m.f.  =  oa;  to  obtain  an  equal  terminal  voltage 
at  full  load  the  field  m.m.f."  must  be  increased  by  an  amount  ad; 
of  this  ac  is  required  to  overcome  armature  reaction  and  cd  is 
required  to  supply  the  armature  resistance  drop. 

In  Fig.  169  the  horizontal  line  through  the  no-load  voltage 
point  m  cuts  the  various  load  saturation  curves  at  e,  f  and  g. 
The  corresponding  values  of  field  m.m.f.  or  field  current  plotted 
on  a  load  base  give  the  curve  af  which  is  the  field  characteristic 
of  the  generator. 


DIRECT-CURRENT  MACHINERY 


191 


The  increase  of  field  current  under  load  is  obtained  by  gradually 
cutting  out  resistance  from  the  field  rheostat  in  series  with  the 
field  winding  (Fig.  170).  This  regulation  of  the  voltage  must 
be  done  by  hand  and  cannot  take  care  of  sudden  changes  of  load. 

150.  Voltage  Characteristic  of  a  Shunt  Generator. — In  the 
shunt  generator,  armature  reaction  and  armature  resistance 
cause  a  decrease  in  terminal  voltage 
under  load,  but  a  third  condition 
must  also  be  taken  into  account. 
The  field  circuit  is  connected  across 
the  terminals  of  the  armature  and  the 
current  in  it  is  proportional  to  the 
terminal  voltage;  thus,  when  the 
terminal  voltage  decreases  due  to 
armature  reaction  and  armature  resistance,  the  field  current  also 
decreases  and  causes  a  further  decrease  in  the  flux;  and  therefore 
the  terminal  voltage  of  a  shunt-excited  generator  is  less  than  it 
would  be  if  the  machine  were  separately  excited  (Fig.  171). 

The  voltage  characteristic  of  a  shunt  generator  may  be  obtained 
from  the  load  saturation  curves,  Fig.  171,  by  drawing  a  straight 
line  om  from  the  origin  through  the  no-load  voltage  point  m. 
The  slope  of  this  line  represents  the  resistance  of  the  shunt  field 


FIG.   170.— Shunt  generator. 


FieM  Current  O      S   ^  Load        Full  Load      1%  Load        2  Load 

FIG.  171. — Voltage  characteristics  of  a  shunt  generator. 

winding  and  its  intersections  with  the  various  saturation  curves 
give  the  terminal  voltages  corresponding  to  the  respective  loads. 
There  are  two  values  of  voltage  corresponding  to  each  value  of 
load.  The  method  of  obtaining  the  voltage  characteristic  is 
shown. 

The  load  current  can  be  increased  by  decreasing  the  resistance 
in  the  load  circuit  until  the  point  M  is  reached.  If  the  load 
resistance  is  further  decreased  the  armature  current  increases 
for  an  instant  and  then  decreases  as  its  m.m.f.  wipes  out  part  of 


192 


ELECTRICAL  ENGINEERING 


the  flux  and  causes  the  generated  voltage  to  decrease.  Finally, 
when  the  load  resistance  is  zero  and  the  generator  is  short-cir- 
cuited the  terminal  voltage  and  the  field  current  become  zero 
and  the  residual  flux  produces  a  small  voltage  which  is  consumed 
in  driving  the  current  os  through  the  armature  resistance. 

The  maximum  current  output  of  a  shunt  generator  is  usually 
many  times  full-load  current  and  can  be  reached  only  with  small 
machines  of  poor  regulation. 

The  internal  voltage  characteristic  can  be  obtained  by  adding 
the  armature  resistance  drops  to  the  terminal  voltages;  that  is, 
by  adding  the  ordinates  of  curves  1  and  2,  Fig.  171,  curve  3  is 
obtained. 

151.  Effect  of  Change  of  Speed  on  the  Voltage  of  a  Shunt 
Generator. — In  Fig.  172 ('a)  oba  is  the  no-load  saturation  curve 
of  a  shunt  generator  at  normal  speed.  The  point  a  is  the  normal 


0    10  20  30  40  SO  60  70  80  90100110120130140150 
Percent  Field  Current 


40    50  60  70  80  20  1Q(| 
Percent  Speed 


FIG.   172. — Variation   of   voltage   of   a   shunt    generator   with   speed. 

voltage  point  and  the  point  6  is  a  point  below  the  knee  of  the 
curve.  The  slope  of  the  line  oa  represents  the  resistance  in  the 
shunt  field  for  normal  operation  and  the  slope  of  ob  represents 
the  resistance  when  operating  with  the  magnetic  circuit  un- 
saturated. obiai  is  a  no-load  saturation  curve  for  90  per  cent, 
of  normal  speed,  each  of  its  ordinates  is  90  per  cent,  of  the  corre- 
sponding ordinate  of  oba. 

From  these  curves  it  is  seen  that  a  drop  of  speed  of  10  per 
cent,  causes  a  drop  of  voltage  pq  =  14  volts  =  14  per  cent,  when 

16  X  100 
the  machine  is  saturated  and  a  drop  p\qi  =  16  volts  = ^ — 

=  19  per  cent,  when  it  is  unsaturated.  Therefore,  a  saturated 
shunt  generator  is  more  stable  than  an  unsaturated  generator, 
that  is,  its  voltage  changes  less  due  to  slight  changes  in  speed. 


DIRECT-CURRENT  MACHINERY 


193 


At  normal  saturation  to  counteract  the  10  per  cent,  drop  in 
speed  the  field  current  must  be  increased  43  per  cent. 

The  saturation  curves  for  speeds  from  80  to  40  per  cent,  of 
normal  are  also  shown.  They  are  intersected  by  the  line  oa 
at  a2,  a3,  etc.,  and  these  points  show  the  no-load  voltages  cor- 
responding to  the  various  speeds. 

The  voltage  speed  curve  at  no  load  is  shown  in  Fig.  172(6). 
The  voltage  becomes  zero  in  this  case  at  40  per  cent,  of  normal 
speed,  since  the  resistance  line  oa  is  tangent  to  the  saturation 
curve  at  this  srjeed  and  so  does  not  cut  it.  The  effect  of  residual 
magnetism  has  been  neglected;  it  tends  to  change  the  curve  as 
indicated  by  the  dotted  lines.  The  line  oa  in  Fig.  172(6),  shows 
the  voltage  speed  curve  at  no  load  for  a  separately  excited 
generator. 


FIG.  173. — Effect  of  change  of  speed 
on  the  voltage  characteristic  of  a  shunt 
generator. 


FIG.  174. — Effect  of  saturation  on 
the  voltage  characteristic  of  a  shunt 
generator. 


In  Fig.  173  are  shown  the  no-load  saturation  curves  for  speeds 
of  100  per  cent.,  90  per  cent,  and  80  per  cent,  of  normal  and  the 
corresponding  voltage  characteristics,  which  may  be  obtained 
as  in  Fig.  171.  The  curves  for  .the  low  speeds  fall  off  very  quickly 
since  the  field  m.m.f.  is  decreased  more  than  the  speed;  the 
maximum  current  outputs  are  also  much  reduced.  The  voltages 
ori,  or2  and  or3  produced  by  the  residual  magnetism  are  di- 
rectly proportional  to  the  speed  as  are  also  the  currents  osi, 
os2  and  os3  produced  by  them. 

15"2.  Effect  of  Saturation  on  the  Voltage  Characteristic  of.  a 
Shunt  Generator. — In  Fig.  174  is  shown  the  no-load  saturation 
curve  ra  of  a  shunt  generator  intersected  by  three  resistance  lines 
oa  corresponding  to  normal  resistance,  06  to  125  per  cent,  of 
normal  and  oc  to  160  per  cent.  The  characteristic  curves  for 

13 


194 


ELECTRICAL  ENGINEERING 


the  lower  saturations  fall  off  very  quickly  since  the  field  m.m.f. 
is  weak  but  they  all  pass  through  a  single  point  s  since  the  voltage 
produced  by  the  residual  flux  is  the  same  for  all.  An  unsaturated 
shunt  generator  does  not  tend  to  maintain  a  constant  terminal 
voltage  under  load,  that  is,  its  voltage  regulation  is  poor. 

153.  Compound  Generator. — Increase  of  field  m.m.f.  under 
load  can  be  obtained  automatically  by  placing  a  series  winding 
on  the  field  poles  in  addition  to  the  shunt  winding.  The  series 
winding  carries  the  load  current. 

If  the  series  winding  is  designed  so  that  the  terminal  voltage 
is  the  same  at  full  load  as  at  no  load  the  generator  is  flat-com- 
pounded, if  the  terminal  voltage  at  full  load  is  higher  than  at  no 
load  the  generator  is  over-compounded. 

In  Fig.  179  are  shown  the  voltage  characteristics  or  regulation 
curves  of  a  generator:  (1)  self -excited;  (2)  separately  excited; 
(3)  flat-compounded;  and  (4)  over-compounded.  If  a  generator 
is  flat-compounded  so  that  it  gives  the  same  voltage  at  full-load 
as  at  no  load  it  will  be  slightly  over-compounded  at  half  load. 

Referring  to  Figs.  167  and  169  for  a  flat-compound  machine 
the  series  winding  at  full  load  must  provide  a  m.m.f.  equal  to 
ad.  The  number  of  turns  per  pole  on  the  series  winding  may  be 

-4- 


FIG.  175. — Compound  generator 
(short  shunt). 


FIG.  176. — Compound  generator 
(long  shunt). 


found  by  dividing  the  required  ampere-turns  ad  by  the  load 
cuj;ren\j\  In  designing  a  series  field  winding  it  is  advisable  to 
put  on  some  extra  ampere-turns  to  make  up  for  any  decrease  in 
voltage  due  to  decrease  in  the  speed  of  the  prime  mover  under 
load.  After  the  machine  is  completed  the  series  excitation  may 
be  corrected  by  connecting  a  shunt  across  the  terminals  of  the 
series  winding  (Figs.  175  and  176).  By  varying  the  resistance 
of  the  shunt  any  required  portion  of  the  current  may  be  taken 
from  the  series  winding.  In  the  case  of  a  generator  supplying 
a  rapidly  fluctuating  load  the  shunt  to  the  series  winding  must 
be  designed  with  its  inductance  in  the  same  ratio  to  the  induce 


DIRECT-CURRENT  MACHINERY 


195 


tance  of  the  series  winding  as  its  resistance  bears  to  the  resist- 
ance of  the  series  winding  in  order  that  the  variable  current  may 
divide  up  in  the  correct  proportions  to  give  the  required 
compounding. 

154.  Voltage    Characteristic    of   a    Compound    Generator. — 
The  voltage  characteristic  for  a  flat-compound  generator  may  be 


O       Field  M.M.F.       a          di        d          d2  O  J«Load          Full  Load     1^  Load 

sa\  ffy  Load  Current 

FIG.  177. — Voltage  characteristic  of  a  flat  compound  generator. 

obtained  from  the  saturation  curves  in  Fig.  177 (a).  At  full 
load  the  series  excitation  is  ad  and  the  terminal  voltage  is  dm2  = 
the  no-load  voltage  am;  at  half  load  the  series  excitation  is  adi  = 

77  and  the  voltage  is  dimi  slightly  greater  than  am:  at  one  and 

A 

one-half  load  the  series  excitation  is  ad%  =  %ad  and  the  terminal 
voltage  d2m3  is  less  than  am.  The  voltage  characteristic  is  shown 
in  Fig.  177(6). 


m3 


E& 


. 

FIG.   178.  —  Voltage  characteristic  of  an  over-compound  generator. 

The  characteristic  for  a  10  per  cent,  over-compound  generator 
may  be  obtained  in  a  similar  way  from  Fig.  178  (a).  At  full 
load  the  series  excitation  is  af  and  the  terminal  voltage  is  /ra2  = 
110  per  cent,  of  the  no-load  voltage  am;  at  half  load  the  series 

excitation  is  a/i  =  -^   and  the  voltage  is  f\m\]  m\  lies  slightly 


196 


ELECTRICAL  ENGINEERING 


The  voltage  charac- 


Compound 

ompound 

^Separately  Excited 
^Shunt  Excited 


above  the  straight  line  joining  m  and  w2. 
teristic  is  shown  in  Fig.  178(6). 

The  increase  of  excitation  for  full  load  of  includes  a  small 
component  which  is  supplied  by  the  shunt  winding,  since  as  the 
voltage  across  it  has  increased  10  per  cent.,  its  current  and  m.m.f. 
have  increased  in  the  same  proportion.  Allowance  should  be 
made  for  this  in  designing  series  windings. 

155.  Short-shunt  and  Long-shunt  Connection. — In  compound- 
wound  generators  the  shunt  winding  may  be  connected  across  the 
armature  terminals,  Fig.  175,  called  short-shunt,  or  outside  of  the 
series  winding,  Fig.  176,  called  long-shunt.  The  characteristic 
curves  are  not  affected  to  a  great  extent  by  the  difference  in  con- 
nection since  with  the  short-shunt  the  voltage  across  the  shunt 
winding  is  higher  than  with  the  long-shunt  by  the  resistance  drop 

in  the  series  winding  and  the 
current  in  the  series  winding 
is  less  by  the  amount  supplied 
to  the  shunt  winding. 

156.  Regulation. — The 
regulation  of  a  separately 
excited  or  shunt-excited  gen- 
erator is  denned  as  the  rise 
in  voltage  when  full  load  is 
taken  off  expressed  as  a  per 
cent,  of  full-load  voltage. 
Fig.  179  shows  the  characteristic  curves  of  a  generator:  (1) 
separately  excited,  (2)  shunt-excited,  (3)  flat-compounded  and 
(4)  over-compounded.  The  per  cent,  voltage  regulation  of  the 

be 

separately  excited  generator  is  — ;.  X  100  per  cent. ;  the  regulation 

CJ 

of  the  shunt  generator  is  -^  X  100  per  cent.  The  regulation  of 
the  flat-compounded  generator  is  the  per  cent,  variation  of  volt- 
age from  the  normal  no-load  or  full-load  value;  it  is  — -  X  100 
per  cent.  The  regulation  of  the  over-compounded  generator  is 
rr  X  100  per  cent.,  where  gh  is  the  maximum  intercept  between 

the  voltage  characteristic  and  the  straight  line  va. 

57.  Series  Generator. — The  series  generator,  Fig.  180,  is 
excited  by  a  series  winding  carrying  the  load  current  and  has  no 


25  50  75 

Per  Cent  Load 


100        125 


FIG.  179. — Voltage  characteristics  of 
generators. 


DIRECT-CURRENT  MACHINERY 


197 


shunt  winding.  Its  characteristic  curves  are  shown  in  Fig.  181. 
The  effect  of  residual  magnetism  is  neglected.  Curve  1  is  a  no- 
load  saturation  curve  obtained  by  exciting  the  generator  from  a 
separate  source.  Curve  2  is  the  internal  voltage  characteristic 
showing  the  relation  between  generated  voltage  and  load  when 
the  generator  is  self-excited, 
and  the  horizontal  distance 
between  curve  1  and  curve  2 
represents  the  effect  of  arma- 
ture reaction.  The  armature 
reaction  due  to  full-load  cur- 
rent is  represented  by  mp,  and 
assuming  that  the  armature 
reaction  varies  directly  as  the 

load  current  the  other  points  on  curve  2  may  be  found  easily. 
Curve  3,  the  external  characteristic,  is  obtained  by  subtracting 
from  the  ordinates  of  curve  2  the  corresponding  resistance 
drops  in  the  armature  and  series  field  shown  in  curve  4. 

The  terminal  voltage  of  a  series  generator  can  be  varied  by  con- 
necting a  variable  resistance  in  shunt  to  the  series  winding.     The 


FIG.   180. — Series  generator. 


100  150 

Percent  Load  Current 


FIG.  181. — Voltage  characteristics  of  a  series  generator. 

slope  of  the  line  oq  represents  the  resistance  in  the  load  circuit 
at  full  load;  as  this  resistance  is  increased  the  load  current  and 
the  terminal  voltage  decrease.  When  the  resistance  line  oC 
is  tangent  to  the  voltage  characteristic  the  machine  is  unstable 
and  is  liable  to  lose  its  voltage.  The  load  resistance  at  which  this 
condition  occurs  is  called  the  critical  resistance  for  the  generator. 


198  ELECTRICAL  ENGINEERING 

It  is  an  indefinite  quantity  on  account  of  the  effect  of  residual 
magnetism. 

158.  Electric  Motors.  —  In  generators  mechanical  power  is 
supplied  and  electrical  power  is  generated.  The  speed  is  fixed  by 
the  prime  mover  and  is  constant.  The  terminal  voltage  is  ap- 
proximately constant  in  the  shunt  generator  and  flat-compound 
generator  and  increases  with  load  in  the  over-compound  generator 
and  the  series  generator.  The  generated  voltage  is  always  greater 
than  the  terminal  voltage  by  the  drop  in  the  armature  resistance  ; 
it  is 

8  =  E  +  Ir. 

In  motors  electrical  power  is  supplied  and  mechanical  power  is 
generated.  The  impressed  voltage  is  fixed  by  the  supply  circuit 
and  is  constant.  The  speed  is  either  approximately  constant  as 
in  the  shunt  motor  or  decreases  with  load  as  in  the  compound 
motor  and  series  motor.  The  voltage  generated  in  the  armature 
has  the  same  equation  as  the  voltage  in  a  generator,  but  it  is  a 
back  voltage  and  opposes  the  current;  the  impressed  voltage  E  is 
greater  than  the  back-generated  voltage  by  the  armature  resist- 
ance drop;  thus, 

,  E  =   &  +  Ir,  (246) 


or  Bt*^      8  =  E  -  Ir.  (247) 


Field  Rheostat      VH,   Rheostat  **.  Field  Rheostat 


FIG.  182. — Shunt         FIG.  183. — Series          FIG.  184. — Corn- 
motor,  motor.  pound  motor. 

159.  Types  of  Motors. — There  are  three  types  of  direct-cur- 
rent motors  corresponding  to  the  three  types  of  generators,  shunt, 
compound  and  series.  The  shunt  motor  has  its  field  circuit  con- 
nected across  the  line  in  shunt  with  the  armature  and  therefore 
has  a,  constant  excitation  (Fig.  182).  The  compound  motor 
has  a  series  winding  carrying  the  load  current  in  addition  to  its 
shunt  winding.  The  excitation  therefore  increases  with  load 
(Fig.  184). 

The  series  motor  has  its  field  circuit  in  series  with  the  armature 
and  has  no  shunt  winding.  Its  excitation  is  zero  at  no  load  and 
increases  directly  with  the  load  current  (Fig.  183). 


DIRECT-CURRENT  MACHINERY 


199 


160.  Speed  Equation  of  a  Motor. — The  impressed  voltage  is 
E  =   8  +  /r, 


and 
where, 

therefore 
and 


8  =  ZnQ    •  10~8  =  Kn$>  volts, 
Pi 


K  =  Z-P-  10~8  is  a  constant; 


E  =  Kn&  +  Ir 

E  -  Ir 

n  = 


(248) 


this  is  the  speed  equation. 

The  term  Ir  is  small  in  comparison  to  E  and  may  be  neglected 
except  at  heavy  loads. 

The  speed  equation  may  therefore  be  written 


n  = 


E 


(249) 


thus,  the  speed  of  a  motor  is  directly  proportional  to  the  impressed 
voltage  and  is  inversely  proportional  to  the  flux  crossing  the  air 
gap. 

161.  Methods  of  Varying  Speed. — The  speed  can  be  varied  in 
three  ways:  (1)  by  varying  the  impressed  voltage  E,  (2)  by  vary- 
ing the  flux  3>  by  reducing  the  field  current,  (3)  by  shifting  the 
brushes. 


•if 


[I        R 


FIG.  185. 


Impressed  Voltagre    E 

FIG.  186. — Variation  of  speed 

with  impressed  voltage. 


1.  The  voltage  impressed  on  the  armature  can  be  varied  by 
introducing  a  resistance  R,  Fig.  185,  in  series  with  the  armature. 
If  Es  is  the  supply  voltage,  the  voltage  impressed  on  the  motor  is 

E  =  E8  -  IR. 


200 


ELECTRICAL  ENGINEERING 


By  increasing  R,  E  can  be  reduced  to  any  required  value  and 
any  speed  from  standstill  to  normal  speed  can  be  obtained  but 
no  increase  of  speed  above  normal. 

Fig.  186  shows  n  as  a  function  of  E;  the  locus  is  a  straight  line 
passing  through  the  origin. 

This  method  of  varying  speed  is  uneconomical  as  a  large 
amount  of  power  is  lost  in  the  control  resistance;  it  is  the  product 
of  the  current  input  and  the  voltage  consumed  in  the  resistance 
and  is 

=  I  X  IR  =  PR  watts. 

The  resistance  R  must  not  be  connected  in  series  with  the  field 
winding  as  it  would  then  decrease  the  field  current  and  therefore 
the  flux  and  tend  to  cause  an  increase  in  speed. 

If  the  speed  is  reduced  to  50  per  cent,  of  normal  in  this  way  the 
efficiency  of  the  motor  is  less  than  50  per  cent. 

With  a  variable  voltage  supply  a  wide  range  of  speed  can  be 
obtained  efficiently  (see  Arts.  173  and  174). 


• 

^-  — 

s* 

^" 

/ 

/ 

j 

/ 

/ 

/ 

1 

Field  Current  \f 

FIG.  187. 

Field  Currant  " 

FIG.  189. 


2.  When  varying  the  speed  by  field  control  the  full  line  voltage 
is  impressed  on  the  armature  and  a  resistance  R  is  connected  in 
series  with  the  field  winding,  Fig.  182.  As  this  resistance  is 
increased  the  field  current  //  decreases  according  to  equation 

E 


and  the  flux  <f>  decreases  with  the  field  current  as  shown  by  the 
saturation  curve  of  the  machine  in  Fig.  187.  Since  the  speed 
varies  inversely  as  3>,  Fig.  188,  the  variation  of  speed  with  field 
current  will  be  represented  by  a  curve  of  the  shape  shown  in 
Fig.  189. 


DIRECT-CURRENT  MACHINERY  201 

By  this  method  the  speed  can  be  increased  to  any  required 
value,  and  it  tends  to  approach  infinity  when  the  field  current  is 
zero  but  is  limited  by  the  residual  magnetism. 

In  machines  of  ordinary  design  the  speed  can  be  increased 
satisfactorily  only  about  70  per  cent,  above  normal  speed  by 
field  weakening.  Beyond  this  point  it  is  not  possible  to  get 
sparkless  commutation  of  full-load  current,  since  the  armature 
m.m.f.  is  strong  enough  to  overcome  the  weak  field  m.m.f.  and 
wipe  out  the  commutating  field  and  at  the  same  time  the  time  of 
commutation  is  decreased. 

By  using  interpoles  which  neutralize  the  effect  of  armature 
m.m.f.  and  provide  a  commutating  field,  the  speed  can  be  in- 
creased to  four  or  five  times  normal  by  field  weakening  without 
injurious  sparking.  • 

The  loss  in  power  in  the  resistance  controlling  the  field  current 
is  very  small  since  the  power  used  for  field  excitation  is  only  1  or  2 
per  cent,  of  the  rated  output. 

3.  At  no  load  with  the  brushes  on  the  neutral  line,  all  the  con- 
ductors on  each  half  of  the  armature  are  effective  in  generating 
the  back  e.m.f.  and  this  is  therefore  the  position  of  minimum 
speed  Fig.  160(a). 

When  the  brushes  are  moved  back  against  the  direction  of 
rotation,  Fig.  160(6),  only  the  belts  of  conductors  under  the  poles 
are  effective  in  generating  the  back  e.m.f.  and  the  result  is  the 
same  as  though  the  flux  had  been  decreased.  The  speed  is  there- 
fore increased. 

Under  load,  when  current  is  flowing  in  the  armature,  the  con- 
ductors between  the  poles  exert  a  demagnetizing  m.m.f.  and  cause 
a  decrease  in  the  flux;  the  speed  therefore  rises  more  than  at  no 
load. 

This  method  of  speed  variation  is  not  used  to  any  extent  since 
it  interferes  with  commutation  and  causes  injurious  sparking. 

162.  Speed  Characteristics  of  Motors.  —  The  speed  charac- 
teristic is  the  curve  showing  the  variation  of  speed  with  armature 
current.  The  speed  equation  was  found  to  be 

n  =  E  ~r  r.p.s.     (Art.  160) 


Shunt  Motor.  —  As  the  motor  is  loaded  and  /  increases  the  speed 
is  affected  in  two  ways:  (1)  the  resistance  drop  Ir  increases  and 
tends  to  cause  a  proportionate  decrease  in  speed;  (2)  the  flux  <£ 


202 


ELECTRICAL  ENGINEERING 


is  decreased  by  the  armature  demagnetizing  and  cross-magnetiz- 
ing m.m.f.;  therefore  the  armature  reaction  tends  to  increase 
the  speed.  The  voltage  drop  at  the  brush  contacts  (Art.  182) 
increases  with  load  but  not  in  direct  proportion.  It  acts  with 
the  armature  resistance  to  decrease  the  speed.  If  the  motor  is 
operated  at  a  point  just  above  the  knee  of  the  saturation  curve 
the  drop  in  speed  due  to  the  armature  resistance  will  be  greater 
than  the  rise  due  to  armature  reaction  and  the  speed  character- 
istic will  fall  as  shown  in  Fig.  190  curve  1. 

The  speed  regulation  of  a  motor  is  the  rise  in  speed  when  full 
load  is  thrown  off  expressed  as  a  per  cent,  of  full-load  speed. 

The  speed  regulation  of  shunt  motors  of  large  size  is  from  2  to 
5  per  cent. 


\ 


1)  Shunt 

2)  Compound 

3)  Differential  Compound 

4)  Series 


(1)  Shunt 

(2)  Series 

(3)  Compound 

(4)  Differential  Compound 


n 


*  %  H          Full 

Load       Load       Load       Load 

Armature  Current  I 


'FiG.  190.  —  Motor  speed  charac- 
teristics. 


H  *  %  Full  I* 

Load  Load  Load  Load          Load 

Armature  Current  I 

FIG.  191. — Motor  torque  charac- 
teristics. 


Compound  Motor. — In  the  compound-wound  motor  the  arma- 
ture resistance  causes  a  drop  in  speed  and  the  m.m.f.  of  the  series 
winding  overcomes  the  effect  of  the  armature  m.m.f.  and  increases 
the  flux  and  thus  decreases  the  speed  more  than  in  the  shunt 
motor.  A  typical  speed  characteristic  of  a  compound  motor  is 
shown  in  curve  2,  Fig.  190. 

If  the  series  winding  is  reversed,  its  m.m.f.  opposes  the  field 
m.m.f.  and  thus  decreases  the  flux  and  causes  the  speed  to  in- 
crease with  load  as  shown  in  curve  3,  Fig.  190. 

The  motor  is  then  called  a  "differential  compound"  motor  and 
may  be  designed  to  give  constant  speed  under  all  loads.  If  the 
series-field  winding  is  strong  the  motor  is  unstable  and  tends  to 
run  at  excessive  speed. 


DIRECT-CURRENT  MACHINERY  203 

Series  Motor.  —  At  no  load  the  series  motor  tends  to  run  at  a 
very  high  speed  limited  only  by  the  residual  magnetism  or  the 
torque  required  to  overcome  the  losses.  As  load  is  applied  the 
current  and  flux  increase  and  the  speed  falls  rapidly  until  the  mag- 
netic circuit  of  the  machine  becomes  saturated;  the  speed  charac- 
teristic then  becomes  almost  horizontal  (Fig.  190,  curve  4). 

163.  Torque  Equation.  —  The  torque  of  a  motor  is  proportional 
to  the  product  of  the  flux  crossing  the  air  gaps  and  the  current 
in  the  armature.  Its  equation  is  derived  as  follows: 

The  voltage  impressed  on  the  armature  is 

E  =  8  +  Ir, 
where 


8  = 

Pi 

is  the  back  voltage  generated  in  the  armature  and  Ir  is  the  voltage 
consumed  by  the  armature  resistance. 
The  power  input  to  the  armature  is 

El  =  87  +  I2r  watts.  (250) 

The  power  lost  in  the  armature  winding  is  I2r  watts,  and  thus 
the  electric  power  transformed  into  mechanical  power  is 

£7  =  ZnQl^lQ-8; 
Pi 

this  is  the  power  output  in  watts  neglecting  the  friction  losses. 
The  motor  output  in  horsepower  is 


746  746 

If  the  torque  developed  is  T  lb.-ft.,  then  the  output  in  horse- 
power is 

0     „  Zn$/^-10~8 

2irnT  Pi 

550  746 
and  the  torque  is 


550  ""~/-  10  8 


T  =  ^ TZfi-  -  =  0.1177Z<i>7^- 10-8  lb.-ft.       (251) 

2irn          74o  pi 

This  is  the  torque  equation  of  a  direct-current  motor. 

T  =  0.1177  X  10-8  X  (Z— )  X(p$)  lb.-ft.,  (252) 


204  ELECTRICAL  ENGINEERING 

where  Z—  is  the  sum  of  all  the  currents  in  the  armature  conduc- 
tors and  p$  is  the  sum  of  the  fluxes  crossing  the  air  gaps  under  all 
the  poles. 

The  torque  equation  may  be  written 

T  =  K3>I  lb.-ft.,  (253) 

where  K  =  0.1177  Z^-  10~8  is  a  constant. 
Pi 

The  torque  of  a  motor  is  therefore  directly  proportional  to  the 
armature  current  and  to  the  flux  in  the  air  gap. 

This  is  the  torque  developed  at  the  shaft  of  the  motor.  The 
torque  output  is  less  due  to  the  iron  and  friction  losses. 

164.  Torque  Characteristics  of  Motors. — The  torque  charac- 
teristic is  a  curve  showing  the  relation  between  the  torque  and 
the  armature  current. 

The  torque  equation  is 

T  =  K&I.     (Art.  163.) 

Shunt  Motor. — In  the  shunt  motor  <i>  is  almost  constant  under 
load,  decreasing  only  by  a  small  percentage  due  to  armature  re- 
action. The  torque  therefore  varies  almost  directly  with  the 
current  (Curve  1,  Fig.  191). 

Series  Motor. — In  the  series  motor  the  flux  increases  almost  in 
direct  proportion  to  the  current,  while  the  magnetic  circuit  is  un- 
saturated,  and  therefore  the  torque  is  proportional  to  the  square 
of  the  current;  at  heavy  load,  when  the  magnetic  circuit  becomes 
saturated,  the  flux  becomes  almost  constant  and  the  torque  then 
increases  in  direct  proportion  to  the  current.  Curve  2,  Fig.  191, 
is  a  typical  torque-current  curve  for  a  series  motor. 

Compound  Motor. — In  the  compound  motor  the  flux  increases 
with  load  but  not  in  direct  proportion  to  the  current ;  thus,  the 
torque-current  curve  (curve  3)  lies  between  those  of  the  shunt 
and  the  series  motor. 

If  the  series  winding  is  reversed,  as  in  the  differential  compound 
motor,  the  flux  decreases  with  load  and  the  torque-current  curve 
falls  below  that  of  the  shunt  motor  (Curve  4). 

165.  Construction  of  the  Speed  Characteristics. — The  speed 
characteristic  of  any  motor  can  be  constructed  if  the  no-load 
saturation  curve  is  given  and  the  magnitudes  of  the  armature 
reaction  and  armature  resistance.     In  Fig.  192  omn  is  the  no- 


DIRECT-CURRENT  MACHINERY 


205 


load  saturation  curve  of  a  motor,  flux  <1>  vs.  field  current  //.  The 
field  current  is  maintained  constant  and  the  flux  at  no-load  is 
$0  =  oa  and  the  speed  is  UQ. 

The  effect  of  armature  reaction  at  full  load  is  represented  by 
mp  and  it  reduces  the  flux  to  the  full-load  value  oai.  At  no  load 
the  impressed  voltage  and  the  back  voltage  are  equal  and  may 
be  represented  by  oa.  The  required  back  voltage  at  full  load 
is  less  than  at  no  load  by  the  armature  resistance  drop  Ir  and 
is  represented  by  oa^  The  speed  at  full  load  is  therefore  reduced 

.     ,,         ..    00,2  ,  ,  oa2 

in  the  ratio  —  to  a  value  n  =  nQ 


'Compound 


Load       Full  Load 

w 


Load 
Current 


FIG.  192. — Construction  of  speed  characteristics. 


If  the  resistance  drop  is  very  small  it  may  happen  that  the 
point  a2  comes  above  ai,  in  this  case  the  speed  rises  above  the  no- 
load  value.  Such  a  condition  often  exists  at  fractional  loads. 
With  overloads  the  demagnetizing  effect  increases  directly  with 
the  current  and  reduces  the  flux  density  considerably  but  on 
this  account  the  cross-magnetizing  effect  is  relatively  smaller 
and  the  increased  resistance  drop  always  results  in  a  decrease 
of  speed.  This  is  the  case  of  the  shunt  machine  and  the  resulting 
speed  characteristic  is  as  shown  in  Fig.  192(c). 

In  the  compound  motor  the  effect  of  armature  reaction  will  be 
opposed  by  the  m.m.f.  of  the  series  winding  and  the  flux  will  be 
reduced  less  than  in  the  shunt  motor  or  it  may  be  increased.  If 
a  series  winding  is  added  to  the  shunt  motor  above  exerting  at 
full  load  a  m.m.f.  rag,  Fig.  192(6),  greater  than  the  armature 
reaction  mp,  the  flux  at  full  load  will  be  increased  to  the  value  oai, 

and  the  speed  will  be  reduced  to  n  =  n0  — 2-     Other  points  on 

the  speed  characteristic  can  be  found  in  the  same  way,  Fig.  192(c). 

166.  Construction  of  the  Speed  Characteristic  for  a  Series 

Motor. — In  Fig.  193  (a)  om  is  the  no-load  saturation  curve  of  a 


206 


ELECTRICAL  ENGINEERING 


series  motor  obtained  by  separately  exciting  the  fields  and  driving 
as  a  generator  at  the  full-load  speed  of  the  motor.  If  mp  repre- 
sents the  effect  of  armature  reaction  at  full  load,  oai  is  the  back 
voltage  at  full  load  at  a  speed  nf  and  it  may  be  taken  to  represent 
also  the  flux  at  full  load.  The  impressed  voltage  is  greater  than 
oai  by  the  resistance  drop  Ir  and  may  be  represented  by  oa. 
This  value  is  constant. 


O  K  Load  Full  Load 

Field  Current  or  Lead  Current 

(a) 


Load  Current 


FIG.  193.  —  Construction  of  speed  characteristics  or  a  series  motor. 


At  one-half  load  the  armature  reaction  may  be  taken  as 


mp 


and  the  corresponding  flux  is  oa3;  the  resistance  drop  is  only 


aa\ 


half  of  its  former  value  =  0,0,2  =  ~?     and  so  the  back  voltage 


must  be  oa2.     The  speed  is  therefore  increased  to  n  =  nf  X 
The  complete  characteristic  is  shown  in  Fig.  193(6). 


100 
90 
80 
70 

r° 

ll° 


^~ 

==>- 

90 
80 

M 

|50 

k 

20 
10 
0 

~7 

/ 

r 

~V 

x 

—  /, 

- 

' 

/ 

•?' 

y 

/ 

—  -^ 

/ 

--,/_ 

1 

^4 

1 

f 

0    10  20  30  40  50   60  70  80  90  100 
Percent  Line  Voltage 


0    10  20  30  40   60  60  70  80    90  100 
Percent  Field  Current  or  Line  Voltage 

(«)  (6) 

FIG.  194. — Variation  of  speed  with  line  voltage  (shunt  motor  without  load). 

167.  Variation  of  Speed  of  a  Shunt  Motor  with  Line  Voltage.— 

If  the  voltage  impressed  on  the  terminals  of  a  shunt  motor 
decreases,  the  speed  decreases  but  not  in  direct  proportion. 
Referring  to  Fig.  194,  (a)  shows  a  no-load  saturation  curve 


DIRECT-CURRENT  MACHINERY  207 

plotted  with  flux  on  a  base  of  field  current  or  line  voltage  and  (b) 
shows  motor  speed  on  a  voltage  base. 

When  the  line  voltage  is  100  per  cent.,  the  field  current,  the 
flux  and  the  speed  are  all  100  per  cent.  If  the  line  voltage  drops 
to  90  per  cent.,  the  field  current  drops  to  90  per  cent,  and  the  flux 
to  97  per  cent;,  due  to  the  decrease  in  voltage  impressed  on  the 
armature  the  speed  tends  to  drop  to  90  per  cent,  but  this  is  partly 
counterbalanced  by  the  drop  in  flux  to  97  per  cent.  The  speed  is, 

therefore,  100  X  0.90  X  JT|=  =  92.9  per  cent.  When  the  volt- 
age falls  to  50  per  cent,  the  flux  is  75.5  per  cent,  and  the  speed  is 
100  X  0.5  X  -  =  62.2  per  "cent. 


If  the  saturation  curve  passes  through  the  origin,  the  speed 
voltage  curve  in  (b)  tends  to  cut  the  speed  axis  about  55.5  per 
cent,  speed.  This  is  a  theoretical  point  for  a  motor  without 
friction  and  without  residual  magnetism. 

If  a  residual  flux  of  10  per  cent,  is  assumed  as  indicated  by  the 
dotted  part  of  the  saturation  curve  the  speed  voltage  curve  drops 
off  to  the  origin  along  the  dotted  part  of  the  curve  in  (b).  The 
motor  becomes  unstable  on  the  lower  voltages. 

168.  Variation  of  Speed  with  Temperature  of  the  Field  Coils.— 
If  the  speed  of  a  motor  is  measured  immediately  after  being 
started  and  again  after  running  under  load  for  some  time  it  will 
be  found  to  have  increased.     If  the  rise  in  temperature  of  the  field 
is  40°C.,  the  increase  in  resistance  will  be  approximately  40  X 
0.4  =  16  per  cent,  and  the  field  current  will  decrease  16  per  cent.; 
this  results  in  a  decrease  of  7  or  8  per'  cent,  in  the  flux  and  an 
increase  of  7  or  8  per  cent,  in  speed.     The  amount  of  the  speed 
change  depends  on  the  saturation  of  the  machine. 

169.  Construction  of  the  Torque  Characteristic.  —  In  Art.  165 
the  method  of  obtaining  the  flux  in  the  air  gap  at  any  given  value 
of  load  current  is  indicated.     If  curves  showing  the  variation  of 
flux  with  load  current  are  plotted  as  shown  in  Fig.  195,  the  torque 
characteristics  of  the  motors  may  be  found  by  multiplying  the 
values  of  flux  by  the  corresponding  values  of  current.     This 
follows  from  the  torque  equation 

T  =  K$I. 

The  torque  at  some  particular  load  must  be  known  in  order  to  fix 
the  scale  of  the  curve. 


208 


ELECTRICAL  ENGINEERING 


The  values  of  torque  obtained  in  this  way  represent  the  torque 
developed  at  the  shaft  of  the  motor.  The  external  torque  or 
available  torque  is  less  by  the  torque  required  to  supply  the  iron 
and  friction  losses. 


Flux=0 


0  %  Load       Full  Lend 

Compound  Motor 

U) 


Load          Full  Load 
ries  Motor 


FIG.  195. — Construction  of  torque  characteristics. 


170.  Starting  Torque.  —  In  Fig.  196  are  shown  typical  speed 
and  torque  characteristics  for  shunt,  series  and  compound  motors 
with  the  same  full-load  speed  and  torque. 

When  starting,  a  motor  must  often  exert  a  torque  considerably 
in  excess  of  the  full-load  value  in  order  to  overcome  friction  and 
the  inertia  of  the  load.  At  the  same  time  the  starting  current 
must  be  kept  within  reasonable  limits  in  order  to  prevent  fluc- 
tuations of  the  voltage  of  the  system.  If  the  required  starting 

torque  is  just  equal  to  full- 
load  torque,  each  type  of 
motor  would  draw  full-load 
current.  If  the  starting  cur- 
rent is  limited  to  150  per  cent. 
of  full-load  current,  the  start- 
ing  torque  developed  by  the 


0     X    X   H 

Load  Current 


0          X   H 

Load  Current 


FIQ.  196.  —  Speed  and  torque 

characteristics. 


,  .  111 

three    motors    would    be  ap- 

.         ,    ,  ,  .  .  _ 

.  proximately,    shunt    140    per 
cent,  of  full-load  torque,  com- 

pound 160  per  cent.,  depending  on  the  strength  of  the  series 
winding  and  series  200  per  cent.  Thus  for  loads  requiring 
starting  torques  in  excess  of  full-load  torque  the  series  and  com- 
pound motors  have  the  advantage  over  the  shunt  motor  that 
they  can  develop  the  required  torque  with  smaller  currents  and 
thus  with  smaller  demands  on  the  supply  system. 

When  starting  shunt  or  compound  motors  the  field  rheostat 


DIRECT-CURRENT  MACHINERY 


209 


Field  Rheostat;; 


FIG. 


Starting  Resistance 

197. — Starting  a  shunt 
motor. 


should  be  entirely  cut  out  in  order  to  give  as  strong  a  field  as 
possible  (Fig.  197). 

171.  Motor  Starter  with  No-voltage  Release. — Fig.  198  shows 
the  connections  of  a  motor  starter  with  a  no-voltage  release. 
When  the  main  switch  is  closed,  current  cannot  flow  to  the  motor 
until  the  starting  handle  is  moved  to  contact  C,  which  allows 

current    to    flow    through    the    field  A 

winding  and  no- voltage  release  B  and 
through  the  total  starting  resistance 
Rs  to  the  motor  armature.  The 
auxiliary  contact  A  is  connected  to 
contact  C  but  is  placed  in  such  a 
position  that,  when  the  motor  is  to  be 
disconnected  the  starting  handle 
opens  the  circuit  at  A  instead  of  at  C  and  the  spark  produced 
by  the  field  discharge  burns  only  the  small  contact  A  which  is 
easily  renewed.  In  starting  up  the  handle  should  be  held  on  C 
for  a  short  time  to  allow  the  field  of  the  motor  to  build  up.  As 
the  motor  speeds  up  and  develops  its  back  voltage  the  handle  is 
gradually  moved  over  to  the  final  position  in  which  all  the  start- 
ing resistance  is  cut  out  and  the 
armature  is  connected  directly 
to  the  supply.  The  starting  re- 
sistance is  cut  into  the  field  cir- 
cuit but  its  effect  in  reducing  the 
field  current  is  usually  negligible. 
An  iron  armature  is  carried  by 
the  starting  handle  and  in  the 
final  position  it  completes  the 
magnetic  circuit  of  the  no- 
voltage  release  and  the  handle 
is  held  in  the  running  position. 
If  the  power  is  interrupted  for 
any  reason,  or  if  the  field  circuit 
opens,  the  no-voltage  release  becomes  demagnetized  and  a 
spring  pulls  the  handle  back  to  the  starting  position. 

The  full  starting  resistance  Rs  must  have  such  a  value  that  the 
starting  current  will  not  be  greater  than  one  and  one-fourth  to 
one  and  one-half  times  full-load  current, 

The  starting  resistance  is  designed  only  for  starting  duty  and 
must  not  be  used  to  operate  the  motor  at  low  speeds  as  it  would 

14 


FIG. 


198. — Direct-current  motor 
starter. 


210  ELECTRICAL  ENGINEERING 

become  overheated  and  burn  out.     It  is  mounted  in  a  ventilated 
box  on  the  back  of  the  starter. 

Example.  —  A  100-hp.,  230-volt,  shunt  motor  has  an  efficiency  of  90  per 
cent,  and  its  armature  resistance  is  r  =  0.03  ohm.  Find  (a)  the  starting 
resistance  to  limit  the  starting  current  to  one  and  one-fourth  times  full-load 
current  and  (6)  the  armature  current  if  the  motor  were  connected  directly 
to  the  supply  while  at  standstill. 

(a)  The  full-load  current  is  030  v  Q  on  =  ^®®  amP-Jtne  allowable  start- 
ing current  is  %  X  360  =  450  amp. 

The  starting  resistance  must  therefore  be 

OOA 

*    R>  =  7^==  0.51  ohm. 

4oU 

This  value  includes  the  small  armature  resistance. 
(6)  If  full  voltage  were  applied  to  the  armature  at  rest  the  current  would  be 
limited  by  the  armature  resistance  only  and  would  tend  to  reach  a  value 

230 
7  =  Q^y5  =  7,600  amp.    This  would  burn  out  the  armature  winding  unless 

the  circuit  was  protected  by  circuit-breakers  or  fuses. 

172.  Adjustable  Speed  Operation.  —  In  Art.  161  three  methods 
of  varying  motor  speeds  were  discussed:  (a)  field  control  for  in- 
creasing the  speed;  (b)  armature  control  for  decreasing  the  speed; 
and  (c)  shifting  of  the  brushes  which  is  possible  only  in  special 
cases. 

For  lathes  where  the  cutting  speed  must  be  constant  although 
the  diameter  of  the  material  changes  field  control  must  be  used. 
b  Armature  con,trol  is  not  satisfactory  as  may 

/^~  ~~^\  be  seen  from  the  following  example.     Refer- 

ring to  Fig.  199,  abc  is  a  piece  of  metal  to  be 
turned.      Assume  that  the  motor  driving  the 
.x  ,  lathe  is  operating  on  a  115-volt  system  and 

that  while  cutting  through  the  projection  be- 
.  199*  tween  a  and  b  the  armature  current  is  100 

amp.,  the  voltage  across  the  armature  terminals  is  75  volts  and 
the  speed  is  300  r.p.m.     This  would  require  a  resistance  in  series 

with  the  armature  R  =  —  r~7  —  ~  0.4  ohms  to  take  up  the  extra 


\a 

k  I  Tool 


40  volts. 

After  passing  the  point  6  the  cut  is  light  and  the  current  is 
reduced  to  10  amp.,  the  drop  across  the  series  resistance  is 
10  X  0.4  =  4  volts  and  the  voltage  impressed  on  the  armature  is 

111;  the  speed  therefore  rises  to  300  X  -       =  444  r.p.m.  and  the 


DIRECT-CURRENT  MACHINERY 


211 


point  of  the  tool  is  liable  to  be  broken  when  it  strikes  the  point 
a  again. 

Field  control  alone  cannot,  however,  give  a  wide  enough 
range  for  all  cases  since  the  speed  can  be  increased  only  about 
70  per  cent,  above  normal  without  trouble  due  to  commutation. 
Where  speed  ranges  of  3  to  1  or  4  to  1  are  required  it  is  necessary 
to  use  interpole  motors  or  to  have  a  multiple-wire  system  of 
supply. 

173.  Multiple-wire  Systems  of  Speed  Control.— In  Fig.  200 
G  is  the  main  generator  giving  250  volts  between  terminals. 


FIG.  200. — Multiple-wire  system. 


FIG.  201. — Three- wire  generator. 


BiBzBz  is  a  balancer  set  with  three  armatures  on  the  same 
shaft  giving  voltages  of  60,  80  and  110.  With  this  system  a  com- 
plete range  of  voltages  from  60  to  250  is  available  for  the  motor 
armature  and  the  intermediate  speeds  can  be  obtained  by  varying 
the  field  of  the  motor  which  is  permanently  connected  across  the 
outer  wires  at  250  volts.  The  possible  speed  range  is  about  6 
to  1,  but  the  system  is  very  complicated  and  expensive. 

Three-wire  systems  are  often  used  instead  of  four- wire  systems. 
That  shown  in  Fig.  201  is  very  common.  A  three- wire  generator 
(Art.  367)  giving  230  volts  .. 


Supply] 


between  outers  and  115  volts 
to  neutral  is  used.  A  speed 
range  of  4  to  1  is  possible  but 
the  motors  must  be  specially 
designed  to  give  a  2  to  1  range 
by  field  control. 

174.  Ward  Leonard  System 
of  Speed  Control. — This  system  is  illustrated  in  Fig.  202.  MI  GI 
is  a  high-speed  motor-generator  set.  The  field  of  the  generator 
is  arranged  so  that  it  can  be  varied  through  a  wide  range  and 
reversed.  Its  terminal  voltage  is  impressed  on  the  armature  of 


Motor  Generator  Set          Adjustible  Speed  Motor 

FIG.  202. — Ward  Leonard  system. 


212  ELECTRICAL  ENGINEERING 

the  motor  M  in  which  speed  control  is  required.  A  uniform 
variation  of  speed  from  a  maximum  in  one  direction  through 
zero  to  a  maximum  in  the  opposite  direction  may  be  obtained. 
The  field  of  the  motor  M  is  permanently  connected  across  the 
supply  lines. 

175.  Speed  Control  of  Series  Motor. — The  speed  of  a  series 
motor  can  be  varied  by  connecting  a  resistance  in  series  with  the 
armature  but  in  cases  where  a  number  of  motors  are  connected 
to  the  same  load,  as  in  electric-railway  operation,  series-parallel 
control  may  be  used  resulting  in  a  more  efficient  speed  variation. 

@    •— VW W^M^+W^  Start 


Speed 


Speed 


0  . —  Full  Speed 

W 

R_ 

FIG.  203.— Series-parallel  control. 

In  Fig.  203  M i  and  M%  are  two  similar  motors  with  their  equal 
starting  resistances  Ri  and  R*.  In  position  (1)  the  whole  of  the 
line  voltage  (500  volts)  is  taken  up  by  the  starting  resistances 
Ri  and  R*  and  the  motors  are  at  rest.  In  position  (2)  the  starting 
resistances  are  short-circuited,  the  current  is  the  same  as  before 
and  each  motor  is  running  at  half  speed  and  develops  a  back 
voltage  of  250  volts.  In  (3)  the  motors  are  in  parallel  and  the 
starting  resistances  are  in  series  again.  The  current  to  each 
motor  is  the  same  as  before  but  the  current  from  the  line  is 
doubled.  The  speed  is  the  same  as  in  (2).  In  (4)  the  resist- 
ances are  short-circuited  again  and  the  motors  are  running  at  full 
speed  with  500  volts  impressed  on  their  terminals.  The  loss  of 
energy  in  the  control  resistances  is  just  half  what  it  would  have 
been  if  the  motors  had  been  connected  in  parallel  at  the  start. 


DIRECT-CURRENT  MACHINERY  213 

The  current  per  motor  is  kept  constant  so  that  the  rate  of  accel- 
eration of  the  car  may  be  constant. 

If  speeds  higher  than  full  speed  are  required  they  may  be 
obtained  by  connecting  resistances  in  shunt  to  the  motor  fields. 

When  it  is  necessary  to  start  a  series  motor  very  slowly  the 
armature  is  shunted  by  a  resistance  which  reduces  the  starting 
torque. 

When  it  is  necessary  to  bring  a  series  motor  to  rest  quickly  it 
may  be  disconnected  from  the  line  and  a  resistance  connected 
across  the  armature  terminals.  The  motor  acts  as  a  generator 
and  slows  up  very  quickly.  The  power  developed  is  wasted  in 
heating  the  resistance.  This  is  one  form  of  dynamic  braking. 

176.  Interpole    Motors. — Interpole    motors    are    fitted    with 
small  poles,  midway  between  the  main  poles,  excited  by  the  load 
current  (Fig.  224) .     The  interpole  m.m.f .  increases  directly  with 
the  load ;  it  is  stronger  than  the  armature  m.m.f.  and  neutralizes 
it  in  the  space  directly  beneath  the  interpole  and  produces  a  flux 
in  the  proper  direction  to  assist  commutation.     The  brushes  are 
placed  on  the  no-load  neutral  and  the  motor  may  be  operated  in 
either  direction.     Since  commutation  is  taken  care  of  by  the 
interpole  field  speed  ranges  of  4  to  1  or  even  6  to  1  can  be  ob- 
tained by  field  control  without  serious  sparking. 

To  prevent  saturation  of  the  interpoles  at  heavy  loads  they 
must  be  designed  with  very  low  flux  densities  at  normal  load. 

If  shunts  to  the  interpole  windings  are  used  the  ratio  of  their 
inductance  to  the  inductance  of  the  interpole  windings  must  be 
the  same  as  the  ratio  of  their  resistances,  in  order  that  sudden 
variations  of  load  may  be  taken  care  of. 

177.  Applications  of  Motors. — When  deciding  on  the  type 
of  motor  to  be  employed  in  a  given  case  three  points  are  of  special 
importance:  (1)  speed — whether  constant,  variable  or  adjustable; 
(2)    starting    torque — whether    greater    or   less    than    full-load 
torque;  (3)  starting  current — whether  it  is  liable  to  disturb  the 
voltage  of  the  system. 

The  types  of  motor  available  are:  (1)  the  shunt  motor  which 
runs  at  approximately  constant  speed  and  develops  a  torque 
proportional  to  the  current;  the  speed  may  be  adjusted  to  any 
suitable  value  within  limits;  (2)  the  series  motor  with  variable 
speed  and  torque  proportional  to  the  square  of  the  current  below 
saturation;  if  twice  full-load  torque  is  required  at  start  it  draws 
about  one  and  one-half  times  full-load  current;  (3)  the  compound 


214  ELECTRICAL  ENGINEERING 

motor  with  variable  speed  and  a  torque  characteristic  between 
the  shunt  and  the  series.  The  compound  motor  has  the  advan- 
tage over  the  series  motor  that  it  approaches  a  limiting  speed  at 
light  load  set  by  its  shunt  excitation. 

Shunt  motors  are  used  for  lathes,  boring  mills,  and  all  constant 
speed  machine  tools,  for  driving  line  shafting  when  the  starting 
load  is  not  too  heavy,  for  fans,  centrifugal  pumps,  etc. 

Series  motors  are  used  in  electric-railway  operation  and  for 
cranes,  hoists,  etc.,  where  very  large  starting  torque  is  necessary 
and  where  constant  speed  is  not  required. 

When  a  load  comes  on  a  series  motor  it  responds  by  decreasing 
its  speed  and  supplying  the  increased  torque  with  a  small  in- 
crease of  current,  thus  preventing  a  sudden  shock  on  the  system. 
A  shunt  motor  under  the  same  conditions  would  hold  its  speed 
nearly  constant  and  would  supply  the  required  torque  with  a 
large  increase  of  current  and  would  thus  make  a  heavy  demand  on 
the  system. 

Series  motors  must  not  be  used  for  belt  drives  or  in  any  case 
where  the  load  may  be  removed  suddenly  since  they  run  at  excess- 
ive speed  at  light  loads.  % 

Compound-wound  motors  are  used  in  classes  of  work  where 
constant  speed  is  not  necessary  and  where  a  fairly  large  starting 
torque  is  required,  except  in  those  cases  in  which  series  motors 
must  be  employed  on  account  of  their  very  large  starting  torque. 

For  express  elevators  a  compound-wound  motor  is  employed, 
the  series  winding  is  required  to  give  the  large  torque  at  start  and 
is  cut  out  after  a  certain  speed  is  reached.  The  motor  then  runs 
at  constant  speed  as  an  ordinary  shunt  machine. 

For  operating  shears,  punches,  etc.,  where  a  high  maximum 
load  must  be  carried  for  a  short  period  a  compound  motor,  with 
a  flywheel  attached,  is  used.  The  drooping  speed  characteristic 
is  necessary  to  enable  the  flywheel  to  give  up  part  of  its  stored 
energy  to  supply  the  peak  of  the  load  and  so  relieve  the  supply 
system.  A  shunt  motor  would  not  drop  in  speed  and  the  fly- 
wheel would  not  be  of  any  value.  A  series  motor  would  not  be 
suitable  as  it  would  run  at  excessive  speeds  before  and  after  the 
cut. 

In  rolling  mills  where  the  load  fluctuates  very  rapidly  a  similar 
compound  motor  with  a  heavy  flywheel  is  used. 

Compound  motors  are  often  installed  in  positions  where  the 
constant  speed  characteristic  of  the  shunt  motor  would  be  more 


DIRECT-CURRENT  MACHINERY  215 

suitable  but  where  the  series  winding  is  required  to  supply  the 
torque  to  overcome  the  inertia  of  the  heavy  moving  parts,  as  in 
the  case  of  heavy  planers,  etc. 

178.  Losses  in  Direct-current  Machinery. — The  main  power 
losses  in  direct-current  generators  and  motors  may  be  divided 
into  copper  losses,  iron  losses  and  friction  losses  and  these  may  be 
subdivided  as  follows: 

Copper  losses : 

Shunt-field  copper  loss. 

Series-field  copper  loss.    * 

Armature  copper  loss. 
Iron  losses: 

Hysteresis  loss. 

Eddy-current  loss. 
Friction  losses: 

Brush-friction  loss. 

Bearing-friction  loss. 

Windage  loss. 

179.  Shunt-field  .Loss. — The  shunt-field  copper  loss  is  7/2r/ 
watts,  where  //  is  the  field  current  and  r/  is  the  resistance  of  the 
winding  at  the  operating  temperature  of  the  machine.     This  loss 
can  be  expressed  as  E/I/,  where  E/  =  J/r/  is  the  voltage  im- 
pressed on  the  field  winding.     All  the  energy  supplied  to  the  field 
winding  is  transformed  into  heat  and  is  wasted,  since  no  energy 
is   required    to    maintain  the  magnetic   flux  after  it  is   once 
established. 

The  shunt-field  loss  is  constant  under  all  conditions  of  load  in 
the  shunt  motor  and  the  flat-compound  generator;  it  decreases 
slightly  with  load  in  the  shunt  generator  and  increases  with  load 
in  the  over-compound  generator. 

The  loss  ranges  from  about  1  per  cent,  of  full-load  output  in 
the  case  of  large  high-speed  machines  to  5  per  cent',  in  small  low- 
speed  machines. 

If  there  is  a  rheostat  connected  in  series  with  the  shunt-field 
winding,  the  power  wasted  in  it  should  be  included  in  the  shunt- 
field  copper  loss. 

180.  Series-field  Loss. — The  series-field   copper  loss  is  I,2rt 
watts,  where  Is  is  the  current  in  the  series  winding  and  ra  is  its 
resistance.     This  loss  increases  as  the  square  of  the  load  current 
of  the  machine.     In  interpole  machines  the  resistance  r,  will  in- 


216 


ELECTRICAL  ENGINEERING 


elude  the  resistance  of  the  interpole  winding.  The  loss  in  the 
shunts  to  the  series  winding  or  in  the  series-field  rheostat  must 
be  included  in  the  series-field  loss. 

181.  Armature  Copper  Loss. — The  armature  copper  loss  pro- 
per is  Ia2ra  watts,  where  Ia  is  the  armature  current  and  ra  is  the 
resistance  of  the  winding  not  including  the  brush  contacts.     It 
increases  as  the  square  of  the  load  current. 

In  addition  to  the  loss  due  to  the  load  current  there  are  extra 
losses  produced  by  local  currents  in  the  armature  coils  short- 
circuited  by  the  brushes.  If  the  e.m.fs.  induced  in  the  various 
parts  of  the  armature  winding  connected  in  multiple  between  the 
terminals  are  not  all  equal  circulating  currents  will  flow  both  at 
no  load  and  under  load  and  cause  loss. 

182.  Loss  at  the  Brush   Contacts. — Closely   associated   with 
the  armature  copper  loss  and  usually  included  with  it  is  the  loss 
at  the  brush  contacts.     Between  the  carbon  brush  and  the  copper 

commutator  there  is  a  drop  of 
voltage  which  is  more  of  the 
nature  of  a  back  voltage  than  a 
voltage  consumed  by  resistance. 
This  drop  varies  with  the  current 
density  and  is  also  affected  by 
the  direction  of  the  current  flow. 
Fig.  204  shows  the  relation  be- 


0.3 


§•0.6 
Q 


© 


(1)  Carbon  to  Copper -Negatire  Brush 

(2)  Copper  to  Carbon -Positire  Brush 


20  30  40 

Amperes  per  Sq.  In. 


FIG.  204. — Brush  contact  drop. 


tween  contact-drop  and  current 
density  for  an  ordinary  hard 
carbon  brush.  The  drop  is 
higher  at  the  negative  brush, 
that  is,  from  carbon  to  copper,  than  at  the  positive  brush.  At 
low  densities  it  increases  with  the  current  but  above  35  or  40 
amp.  per  square  inch  it  becomes  almost  constant. 

The  loss  of  power  at  the  brush  contacts  is  ela,  where  e  is  the 
drop  of  voltage  at  the  positive  plus  the  negative  brush  and  may 
be  taken  as  approximately  2  volts  except  for  very  soft  carbon  or 
graphite  brushes  or  for  the  combined  metal  and  carbon  brushes, 
where  it  is  very  much  lower. 

With  a  current  density  of  35  amp.  per  square  inch  and  a  con- 
tact drop  of  1  volt  per  brush  the  power  loss  is  35  watts  per  square 
inch  of  brush  contact.     This  is  a  reasonable  value.     Where  the 
contact  drops  are  smaller  higher  current  densities  may  be  used. 
In  low-voltage  machines  this  loss  of  power  is  an  appreciable 


DIRECT-CURRENT  MACHINERY 


217 


per  cent,  of  the  output  and  may  be  very  serious,  but  for  high- 
voltage  machines  it  is  less  important. 

With  copper  brushes  the  contact  drop  is  very  small  and  densi- 
ties up  to  150  amp.  per  square  inch  are  used. 

183.  Hysteresis  Loss.  —  The  hysteresis  loss  is  due  to  the  rever- 
sal of  the  magnetic  flux  in  the  armature  iron  as  it  moves  across  a 
pair  of  poles. 

The  loss  per  cubic  centimeter  of  iron  per  cycle  is 

WH  =  TjCB1-6  ergs     (equation  123), 

where  (B  is  the  maximum  induction  density  in  lines  per  square 
centimeter  and  TJ  is  the  hysferetic  constant  for  the  iron.  The 
value  of  tj  for  ordinary  armature  iron  after  assembly  may  be 
taken  as  0.0027. 

The  number  of  cycles  of  magnetism  per  second  or  the  frequency 
is 


7} 

f  =  r.p.s.  X  number  of  pairs  of  pole  s  =  n  -• 


(254) 


If  the  induction  density  B  is  expressed  in  lines  per  square  inch, 
the  hysteresis  loss  per  cubic  inch  of  iron  per  second  is 

—  2)     (2.54)3/  ergs  per  second 
.o4  / 

/  10~7  watts.  (255) 


Wh  = 


FIG.  205. — Distribution  of  flux  in  the  armature  core. 

In  transformers  where  the  induction  density  is  very  nearly  uni- 
form throughout  the  volume  of  the  iron,  this -value  multiplied 
by  the  volume  of  the  iron  would  give  the  hysteresis  loss  very 
closely;  but  in  the  case  of  dynamos  the  induction  density  is  not 


218  ELECTRICAL  ENGINEERING 

uniform  but  varies  from  a  maximum  at  the  roots  of  the  teeth 
to  a  very  low  value  at  the  bottom  of  the  armature  core,  as  indi- 
cated in  Fig.  205.  The  loss  can  only  be  determined  very  approxi- 
mately by  making  separate  calculations  for  the  teeth  and  the  core. 
The  hysteresis  loss  varies  directly  with  the  frequency,  that  is, 
with  the  speed  of  the  machine;  it  increases  as  the  1.6th  power  of 
the  flux  density  and  thus  of  the  voltage  and  it  increases  to  some 
extent  under  load  due  to  the  distortion  of  the  flux.  The  loss  is 
increased  in  sections  where  the  density  is  increased  more  than  it 
is  decreased  where  the  density  is  decreased. 

The  hysteresis  loss  may  be  very  seriously  increased  by  careless 
handling  of  the  materials  during  assembly. 

In  some  cases  the  hysteresis  loss  has  been  found  to  increase 
after  the  machine  has  been  in  operation  for  a  short  time.  This  is 
due  to  aging  of  the  iron  and  will  increase  the  temperature  rise  of 
the  machine.  Silicon  steel  in  additio-n  to  having  a  low  hysteresis 
loss  and  high  electrical  resistance  is  non-aging. 

184.  Eddy-current  Loss.  —  The  eddy-current  loss  is  due   to 

electric    currents    set    up    in    the 
armature  iron  by  the  e.m.fs.  gener- 
ated in  it  as  it  cuts  across  the  flux. 
In  Fig.  206  abed  represents  a  sec- 
tion of  an  armature  punching  of 
thickness  t  in.     If  the  flux  density 
in  the  gap  is  B  lines  per  square 
inch  and  the  edge  ab  is  moving  with 
FIG.  206.—  Eddy-current  loss  in     a  velocity  of  v  in.  per  second  across 

the  armature  core. 

the  flux,  then,  the  e.m.f.  generated 
in  the  length  ab  is 

e  =  BtvlQ'8  volts. 

This  e.m.f.  will  cause  a  current  to  circulate  through  the  iron  as 
indicated  by  the  arrows  and  its  value  will  be 

i  =  kye  =  kyBtvlO'8  amp. 

where  7  is  the  electric  conductivity  of  the  iron  and  k  is  a  constant 
depending  on  the  dimensions  of  the  section. 
The  loss  in  the  section  will  be 


p  =  i*r  =  J     =  10-i6  =  fclT£W  watts  (256) 

KJ  Icy 

where  ki  =  k  X  10~16  is  a  constant. 


DIRECT-CURRENT  MACHINERY 


219 


The  eddy-current  loss,  therefore,  varies  as  the  square  of  the 
induction  density,  that  is,  the  square  of  the  terminal  voltage, 
the  square  of  the  thickness  of  the  plates,  and  the  square  of  the 
speed;  it  also  depends  on  the  conductivity  of  the  iron  used.  It 
cannot  be  calculated  accurately  in  the  case  of  an  armature  where 
the  induction  density  varies  throughout  the  volume. 

The  loss  increases  under  load  due  to  the  distortion  of  the  field 
but  it  tends  to  decrease  as  the  temperature  rises  and  decreases 
the  conductivity. 

The  eddy-current  loss  is  reduced  by  building  up  the  armature  of 
thin  punchings  insulated  from  one  another  by  varnish  which  in- 
creases the  resistance  in  the  path  of  the  eddy  currents.  If  the 
slots  are  filed  out  after  assembly  to  remove  rough  edges,  adjacent 
plates  may  be  short-circuited  and  the  value  of  the  lamination 
of  the  core  partly  lost.  If  the  punchings  are  assembled  under  too 
great  pressure  the  resistance  between  them  will  be  reduced  and 
the  loss  increased. 


I      I 


i     1 

LJ 


a 

in 


CA; 


FIG.  207. — Eddy-current  loss  in  the  pole  face. 

Eddy-current  losses  also  occur  in  the  armature  end  plates, 
spider  and  bolts,  due  to  leakage  fluxes,  and  they  are  comparatively 
large  since  these  parts  are  not  laminated. 

185.  Pole-face  Loss. — Eddy  currents  are  also  produced  in 
the  pole  faces  due  to  local  variations  of  the  induction  density  as 
the  armature  teeth  move  across  them  (Fig.  207).  The  pulsations 


220  ELECTRICAL  ENGINEERING 

of  flux  do  not  extend  far  below  the  surface  of  the  pole  face  since 
they  are  opposed  by  the  eddy  currents.  The  frequency  is  very 
high,  being  proportional  to  the  product  of  the  revolutions  per 
second  and  the  number  of  teeth.  Formulae  have  been  derived 
for  the  calculation  of  the  pole  face  losses  but  they  are  very  com- 
plicated and  must  be  applied  with  great  care. 

To  reduce  the  loss  the  pole  faces  of  all  direct-current  machines 
should  be  laminated.  Machines  with  long  air  gaps  have  small 
pole-face  losses. 

186.  Brush -friction  Loss. — The   brush-friction   loss   in   foot- 
pounds per  second  is  equal  to  the  product  of  the  total  brush 
pressure  in  pounds,  the  peripheral  speed  of  the  commutator  in 
feet  per  second  and  the  coefficient  of  friction  between  the  brush 
and  the  commutator. 

With  carbon  brushes  the  pressure  should  be  from  1.5  to  2 
Ib.  per  square  inch;  this  value  multiplied  by  the  area  of  all  the 
brushes  gives  the  total  brush  pressure  in  pounds.  In  railway 
motors,  where  there  is  a  great  deal  of  vibration,  pressures  up  to 
5  Ib.  per  square  inch  are  used  in  order  to  insure  good  contact. 

The  coefficient  of  friction  between  an  ordinary  hard  carbon 
brush  and  the  commutator  may  be  taken  as  0.3,  for  graphite 
brushes  it  is  about  0.25  and  for  copper  brushes  0.2. 

The  brush  friction  loss  varies  directly  with  the  speed  but  is 
independent  of  the  load. 

187.  Bearing-friction  and  Windage  Losses. — These  two  losses 
are  difficult  to  separate  and  only  their  combined  value  can  be 
obtained  by  test. 

The  bearing-friction  loss  is  proportional  to  the  (speed)  '*  for 
bearing  velocities  up  to  2,000  ft.  per  minute  and  is  approximately 
proportional  to  the  speed  for  higher  values.  It  can  be  calculated 
approximately. 

The  windage  loss  varies  as  the  (speed) 3  but  is  usually  small  up 
to  peripheral  speeds  of  6,000  ft.  per  minute.  It  depends  on  the 
construction  of  the  machine  and  cannot  be  calculated  with  any 
degree  of  accuracy. 

The  combined  friction  and  windage  losses  vary  from  about  H 
per  cent,  of  the  rated  output  in  large  slow-speed  machines  to  2 
or  3  per  cent,  in  small  high-speed  machines. 

In  very  high-speed  machines  the  windage  loss  may  become  the 
larger  part  of  the  combined  loss. 

The  friction  and  windage  losses  are  independent  of  the  load. 


DIRECT-CURRENT  MACHINERY 


221 


188.  Eddy-current    Losses    in    the    Armature    Copper. — In 

addition  to  the  losses  discussed  above  under  certain  conditions 
large  eddy-current  losses  may  occur  in  the  armature  copper  both 
at  no  load  and  under  load.  There  are  two  causes  of  this  loss. 
1.  Some  of  the  flux  from  the  field  poles  passes  down  into  the 
slots  and  cuts  the  armature  conductors  inducing  eddy  currents 
in  them  in  the  directions  indicated  in  Fig.  208(1).  The  propor- 
tion of  the  flux  entering  the  slots  depends  on  the  length  of  the  air 
gap,  the  width  of  the  slots  and  the  saturation  of  the  teeth. 
In  direct-current  machines  of  moderate  size  the  resultant  loss  is 
small.  In  turbo  alternators  with  very  wide  slots  and  long  air 


(i)  (2) 

FIG.  208. — Eddy-current  losses  in  armature  conductors. 

gaps  the  loss  tends  to  be  large  and  the  conductors  in  the  upper 
part  of  the  slot  should  be  laminated  vertically.  Under  load  the 
loss  tends  to  increase,  due  to  the  increased  saturation  of  some  of 
the  teeth. 

2.  A  second  and  more  important  cause  of  eddy  currents  in 
the  copper  is  due  to  the  unequal  distribution  of  flux  in  the  arma- 
ture teeth  near  the  pole  tips  Fig.  208(2).  The  flux  density  in 
tooth  a  is  much  lower  than  that  in  6;  the  magnetic  potential  of 
the  roots  of  the  two  is  the  same,  but  since  a  larger  m.m.f.  is 
required  for  b  than  for  a  the  magnetic  potential  at  the  top  of  6 
must  be  greater  than  that  at  the  top  of  a  and  flux  will  pass  across 
the  slot  as  indicated.  The  flux  density  and  the  e.m.f.  induced  in 
the  conductor  at  the  top  of  the  slot  will  be  greater  than  at  the 
bottom  and  eddy  currents  will  flow  in  the  direction  of  the  e.m.f. 
in  the  upper  part  of  the  conductor  returning  in  the  lower  part. 
The  loss  due  to  these  currents  is  especially  large  in  machines  with 
deep  conductors.  It  is  a  function  of  the  tooth  saturation  and 
therefore  increases  very  rapidly  with  increase  of  voltage;  it 
increases  with  load  due  to  field  distortion  and  also  increases  with 
speed.  In  cases  where  it  is  liable  to  be  large  conductors  should 
be  laminated  in  a  horizontal  plane. 


222  ELECTRICAL  ENGINEERING 

Three  laminated  conductors  are  shown  in -Fig.  208. 
These  eddy-current  losses  in  the  copper  cannot  be  calculated 
but  may  be  kept  small  by  proper  design. 

189.  Constant  Losses  and  Variable  Losses. — The  losses  are 
sometimes  divided  into  two  groups,  the  constant  losses  and  the 
variable  losses. 

The  constant  losses  are  those  which  do  not  vary  to  any  great 
extent  under  load  and  include  the  shunt-field  copper  loss,  the 
iron  losses  and  the  friction  and  windage  losses. 

The  variable  losses  are  those  which  increase  with  load,  namely, 
the  armature  copper  loss  and  the  series-field  copper  loss. 

190.  Core  Loss. — The  core  loss  includes  all  the  losses  located 
in  the  armature  core  except  the  armature  copper  loss.     It  may 
be  obtained  by  running  the  machine  without  load,  measuring  the 
input  and  subtracting  the  shunt-field  copper  loss  and  the  fric- 
tion and  windage  losses.     It  therefore  includes  the  no-load  iron 
losses  and  in  addition  the  eddy-current  losses  in  the  pole  faces 
and  in  the  armature  conductors  and  any  losses  which  may  be 
produced  by  currents  circulating  in  the  armature  windings. 

The  core  loss  increases  to  some  extent  under  load  due  to  field 
distortion. 

In  designing  machines  an  approximate  value  for  the  core  losses 
may  be  obtained  from  curves  such  as  those  in  Fig.  257,  compiled 
from  tests  on  standard  machines. 

191.  Efficiency. — The   efficiency  of  a  machine   may  be  va- 
riously expressed,  as, 

output  _, 

n  =  - — ~~  100  per  cent, 
input 

output 

=  -         .    ,    , —   -  100  per  cent, 
output  +  losses 

input  —  losses  .,__  /OCT\ 

=  -  -  100  per  cent.  (257) 

input 

The  efficiency  varies  with  the  output;  at  light  loads  it  is  low  on 
account  of  the  constant  losses;  between  three-fourths  load  and 
full  load  it  is  maximum  and  the  constant  losses  and  variable  losses 
are  nearly  equal;  above  full  load  it  decreases  again  due  to  the 
rapid  increase  of  the  variable  losses.  The  limit  of  the  efficiency 
which  can  be  reached  commercially  depends  on  the  output,  the 
voltage  and  the  speed.  A  higher  efficiency  can  be  obtained  with 
large  machines  than  with  small  machines.  A  higher  efficiency 


DIRECT-CURRENT  MACHINERY 


223 


can  be  obtained  with  high-voltage  or  high-speed  machines  than 
with  low-voltage  or  low-speed  machines. 

For  220-volt  direct-current  motors  the  full-load  efficiency 
ranges  from  about  85  per  cent,  for  small  sizes  to  93  per  cent,  for 
large  sizes. 


15000 
14000 


0    40    80  120  100  200  240  280  320  360  400 
Amperes  Output 

FIG.  209. — Characteristic  curves  of  a 
200  kw.  compound-wound  generator. 


20  40  60  80    120   160 
Amperes 


200 


FIG.  210. — Characteristic  curves 
of  a  75-h.p.  railway  motor. 


/ 

/ 

/ 

/ 

\ 

/ 

/ 

£              90 

\ 

E 

§ 

ffic 

en 

—  — 

cy 
7 

«^ 

—  — 

^ 

—  —  . 

g  800  >>80 
5           H  70 

\ 

t 

/ 

f  • 

fr 

^ 

7 

,<?. 

Ron   .  48  ^ 

|  600  "60 

4 

,7 

/ 

~~~- 

^—- 

"—  ^, 

600  js    4S   „ 

_1       g 

«?/ 

V 

y 

400  S    32  (5 

1 

0  / 

' 

1         *10 

/ 

0         0 

jr\ 

0             0 

) 

4 

0 

£ 

0 

I 

0 

1C 

0 

2( 

»0 

FIG.  211. — Characteristic  curves 
of  a  500-volt  crane  motor  with  a 
capacity  of  65  h.p.  for  H  hour. 


0    10  20  30  40  50  60  70  80  90  100  110 
Line  Amperes 

FIG.  212. — Characteristic  curves  of  a 
25-h.p.  compound-wound  motor. 


For  550-volt  direct-current  generators  the  full-load  efficiency 
ranges  from  about  90  per  cent,  for  small  sizes  to  96  per  cent,  for 
large  sizes. 

Fig.  209  shows  the  characteristic  curves  of  a  200-kw.  com- 
pound-wound generator,  550  to  625  volts. 

Fig.  210  shows  the  characteristic  curves  of  a  75-h.p.  600-volt 
railway  motor  built  by  the  General  Electric  Co. 


224 


ELECTRICAL  ENGINEERING 


Rotation 


Fig.  211  shows  the  characteristic  curves  of  a  500-volt  crane 
motor  built  by  the  Crocker  Wheeler  Co.  Its  rated  output  is  65 
hp.  for  J^  hr.  with  a  temperature  rise  of  40°  C. 

Fig.  212  shows  the  characteristic  curves  of  a  25-hp.,  500-volt 
compound-  wound  motor  of  the  Westinghouse  Co. 

192.  Commutation.  —  Commutation  is  the  most  serious  limita- 
tion encountered  in  direct-current  machinery. 

Fig.  213  represents  the 
armature  winding  of  a  bipolar 
generator.  The  current  enter- 
ing by  the  brush  B2  divides 
into  two  equal  parts  Ic  which 
follow  the  two  paths  through 
the  winding  and  unite  again  at 
the  brush  BI.  Any  coil  c 
while  moving  from  B%  to  BI 
carries  a  current  Ic.  After 
passing  BI  it  carries  an  equal 
current  Ic  but  in  the  opposite 
direction  and,  therefore,  while 

passing  under  the  brush  B1  the  current  changes  from  Ic  to  -Ic, 
that  is,  it  is  commutated  or  reversed. 

The  factors  to  be  considered  in  a  study  of  commutation  are: 

1.  /c,  the  intensity  of  the  current  to  be  commutated. 

2.  T,  the  time  of  commutation. 

3.  rc,  the  resistance  of  the   contact  of  the  brush  with  the 
commutator. 

4.  r,  the  resistance  of  the  short-circuited  coil. 

5.  L,  the  inductance  of  the  coil. 

6.  The  direction  and  intensity  of  the  flux  cut  by  the  coil  during 
commutation. 

7.  Short-circuit  currents  in  the  coil  when  the  brush  short- 
circuits  an  active  e.m.f. 

8.  Current  density  and  loss  of  energy  at  the  brush  contacts. 
The  current  to  be  commutated  is  that  carried  by  each  con- 

ductor of  the  armature  winding.  If  /  is  the  load  current  of  the 
machine  and  pi  is  the  number  of  paths  in  parallel  through  the 
winding,  the  current  per  conductor  is 


FIG.  213. — Commutation. 


and  increases  directly  as  the  load  current. 


DIRECT-CURRENT  MACHINERY 


225 


The  time  of  commutation  is  the  time  during  which  two  adjacent 
commutator  bars  are  short  circuited  by  the  brush.  In  Fig.  214 
commutation  of  the  current  in  coil  c  begins  as  soon  as  the  brush 
touches  bar  2  and  must  be  completed  when  the  brush  breaks  con- 
tact with  bar  1.  If  the  width  of  the  brush  is  d  in.,  the  thickness 
of  insulation  between  bars  is  5  in.,  and  the  peripheral  speed  of  the 
commutator  is  V  in.  per  second,  the  time  of  commutation  is 


T  = 


d  -  d 
V 


sec. 


Since  5  is  very  small,  the  time  of  commutation  varies  directly  as 
the  width  of  the  brush  and  inversely  as  the  speed  of  the  machine. 
The  time  of  commutation  varies  from  0.0005  to  0.002  sec. 


<z 

nn 

c 

IT 

b 

I 

i  v 

- 

w 

M 

>lc      I 

\       3       1       2      J       1        |                | 

•**'       ]S 

V 

d 

TP 

C 

nh 

b 

TP 

.       Ic' 

,t     i' 

Ic      - 

0 

i        Ic 

0 

r  L 

0 

Ic      , 

!L^bi 


I       3      Pl^sJUj 


2IC 

FIG.  214. 


2IC 
FIG.  215. 


8*0 

FIG.  216. 


The  resistance  of  the  brush  contact  plays  a  very  important  part 
in  commutation;  it  tends  to  reduce  the  current  in  the  short-cir- 
cuited coil  to  zero  and  then  to  build  it  up  in  the  opposite  direction. 
It  would  produce  complete  commutation  if  it  were  not  opposed 
by  the  effects  of  the  resistance  and  inductance  of  the  coil.  Its 
function  is  illustrated  in  Fig.  215.  If  the  resistance  of  the  total 
brush  contact  is  rc,  then  in  Fig.  214  the  drop  of  potential  between 
the  brush  BI  and  bar  1  is  2lcrc.  As  soon  as  the  brush  touches  bar 
2  commutation  begins  and  the  brush-contact  resistance  must  be 
separated  into  two  parts,  r\  the  resistance  from  the  brush  to  bar  1 
and  r2  the  resistance  from  the  brush  to  bar  2.  If  at  the  instant 
represented  in  Fig.  215  the  current  in  the  coil  is  i,  then  the  current 
flowing  from  bar  1  to  the  brush  is  Ic  +  i  and  the  drop  of  potential 
is  (Ie  -[-  i)rimj  the  current  from  bar  2  to  the  brush  is  Ic  —  i  and 
the  drop  of  potential  is  (Ie  —  i)r%.  Since  the  resistance  ri  is  in- 
creasing while  r2  is  decreasing,  the  current  from  bar  2  will  increase 
while  that  from  bar  1  will  decrease  and  the  current  in  the  coil  will 
decrease.  Neglecting  the  resistance  and  inductance  of  the  coil 

15 


226 


ELECTRICAL  ENGINEERING 


the  current  flowing  in  the  coil  will  be  zero  when  ri  and  r2  are  equal 
and  when  therefore  half  of  the  time  of  commutation  has  passed. 
Any  further  increase  in  the  resistance  ri  will  cause  part  of  the 
current  from  coil  6  to  flow  through  coil  c  in  order  to  reach  the 
brush  BI  by  the  path  of  least  resistance.  As  the  resistance  7*1 
still  increases,  more  and  more  current  flows  through  c  until  ri 
becomes  infinite  as  the  brush  breaks  contact  with  bar  1  and  the 
total  current  Ic  from  b  flows  through  c.  Commutation  is  then 
complete. 

In  Fig.  217,  curve  (1),  the  current  in  coil  c  is  plotted  on  a  time 
base  for  half  of  one  revolution;  it  is  reversed  in  the  time  T,  rep- 


[Time  of  Commutation 


(1)  Commutation  by  Contact 
Kesistauce  only 

(2)  Effect  of  the  Besistance 
of  the  Coil 

(3)  Effect  of  the  Inductance 
Of  the  Coil 

(4)  Effect  of  a  Negative 
Commntating  E.M.F. 

(5)  Over-Commutatioa 


Time 


resented  by  OT,  during  which  the  coil  moves  across  the  brush  B\, 
and  it  must  vary  according  to  a  straight  line  law.  This  can  be 
proved  as  follows: 

If  Fig.  215  represents  the  condition  t  sec.  after  the  beginning 
of  commutation,  neglecting  the  resistance  and  inductance  of 
the  coil,  the  drop  of  potential  from  the  commutator  to  the  brush 
must  be  the  same  at  both  sides,  or 


=  rc 


but 

therefore,          /j     ,   ^  r 
Solving  for  i  this  gives 


T  -  t 
T 


and 


T  -  2t 


T 
t 
T 


which  is  the  equation  of  a  straight  line. 

T 
When  t  =  -,i  =  0,  and  when  t  =  T,  i  =  -  Ic. 

4 


(258) 
(259) 


DIRECT-CURRENT  MACHINERY 


227 


If  the  resistance  of  the  coil  is  taken  into  account,  the  drop  of 
potential  across  TZ  (Fig.  216)  must  be  greater  than  the  drop 
across  ri  by  the  amount  required  to  maintain  the  current  i  through 
the  resistance  r;  therefore, 


(Ic  + 


and 


r  (Tt  -  t2)  +  rcT2 


(260) 


T 


When  t  =     ,  i  =  0,  and  when  t  =  T,  i  =  -  Ic. 


The  current  therefore  passes  through  zero  at  the  same  instant 
as  before  and  is  completely  reversed  in  the  same  time,  but  the 
variation  does  not  follow  a  straight  line  law  but  a  curve  as  shown 
in  curve  (2),  Fig.  217.  The  effect  of  the  coil  resistance  is  very 
small  and  may  be  neglected. 

The  effect  of  the  inductance  of  the  armature  coil  must  next 
be  considered.  Armature  coils  are  partially  surrounded  by  iron 
and  therefore  have  a  large  inductance  which  is  proportional 
to  the  square  of  the  number  of  turns  in  the  coil.  With  full-pitch 


(ft) 


Full  Pitch  Winding 

(a) 

FIG.  218. — Inductive  flux  in  a 
full-pitch  winding. 


I  '  —  02 

Short  Pitch  Winding 

(a) 


FIG.  219. — Inductive  flux  in  a 
short-pitch  winding. 


drum  windings,  Fig.  218,  both  the  coils  in  one  slot  will  be  short- 
circuited  at  one  time  and  the  inductive  flux  linking  the  slot  part 
of  each  of  them  will  be  almost  twice  as  large  as  in  the  case  of 
fractional  pitch  windings  (Fig.  219).  The  flux  around  the  end 
connections  is  approximately  the  same  in  the  two  cases.  The 
inductance  of  the  coil  is,  therefore,  partly  self-inductance  and 
partly  mutual  inductance  between  adjacent  coils  but  the  resultant 
effect  may  be  considered  as  due  to  an  inductance  L, 


228  ELECTRICAL  ENGINEERING 

When  the  current  in  a  coil  of  inductance  L  henrys  is  changing 
at  the  rate  -r  amp.  per  second,  an  e.m.f.  L  -j  volts  is  generated 

in  a  direction  opposing  the  change  of  current. 

In  Fig.  216  the  drop  of  potential  from  bar  2  to  the  brush  is  the 
same  by  the  two  paths,  one  through  the  resistance  r2  and  the 
other  through  the  coil  in  series  with  the  resistance  n;  and  thus 

(Ic  +  i)  ri  +  ri  +  L  ^  =  (Ic  -  i)  r2, 
or,  substituting  the  values  of  ri  and  r2  found  above, 

T  rt'i  T 

(I.  +  i)  rc  -—  +  ri  +  L       =  (/.  -  i)  rc  ±-         (261) 


The  complete  solution  of  this  equation  is  difficult  but  a  partial 
solution,  for  the  instant  when  commutation  is  complete,  may  be 
obtained  easily. 

Equation  (261)  may  be  written, 

T  di  ™  ilc  ~  i\  rn  /Ic  +  i\ 

Ldt  =  -n  +  rcT  rid  "  rcT  (f^i)  • 

when  t  =  T  and  i  =  —  Ic 

Ld~t  =  rlc  +  27crc  -  rcT  X  £;  (262) 

this  last  term  is  indefinite  but  its  value  may  be  found  by  differ- 
entiating the  numerator  and  denominator  with  respect  to  the 
independent  variable  t. 

di 
r/c  -f  a"!  di  di 


r  - 

and  substituting  this  in  equation  (262), 

T^  =  rIc  +  2Icrc  +  rcT^ 
dt  at 

or 

di  =  Ic(r  +  2rc) 
dt         L  -  rcT  ' 

If  L  =  rcT,  ~r  =  a  ,  the  reactance  voltage  L  ^  =  oc  ,  and  spark- 

ing will  occur. 

di 
If  L>rcT,  -r  is  positive  and  the  current  tends  to  increase  in- 

stead of  reversing  and  sparking  results. 


DIRECT-CURRENT  MACHINERY 


229 


If  L<rcT,  -T  is  negative  and  finite  and  commutation  may  be 

sparkless  and  satisfactory. 

This  condition  for  sparkless  commutation  may  be  expressed 
in  another  way. 

27 

L  must  be  <  rcT  or  -frL  must  be  <  27>c; 


27  27 

~f£  is  the  average  rate  of  change  of  current  and  -jjfr  L  is  called  the 

average  reactance  voltage,  while  2lcrc  is  the  voltage  drop  at  one 
brush  contact. 

Therefore,  the  condition  for  sparkless  commutation  is  that  the 
average  reactance  voltage  must  be  less  than  the  voltage  drop  at 
one  brush  contact.  This,  however,  applies  only  to  the  case  in 
which  no  flux  is  cut  by  the  short-circuited  coil  and  where  there- 
fore no  e.m.f.  due  to  rotation  is  generated  in  the  coil  either  assist- 
ing or  opposing  commutation.  The  brush-contact  resistance  is 
the  only  factor  operating  to  cause  the  current  to  reverse. 

Due  to  the  effect  of  inductance,  the  current  does  not  decrease 
so  quickly  as  in  curve  1,  Fig.  217,  but  follows  curve  3,  and  the 
rate  of  change  of  current  becomes  very  rapid  as  the  brush  breaks 
contact  with  bar  1  and  sparking  is  liable  to  occur.  The  current 
density  in  the  brush  tip  also  becomes  very  high,  tending  to  burn 
both  the  brush  and  the  commutator. 


No-Load  Neutral 


Load  Neutral 


Load  Neutral 


FIG.  220. — Commutating  field. 

When  the  inductance  and  the  reactance  voltage  are  large  it  is 
necessary  to  have  an  e.m.f.  generated  in  the  coil  to  assist  commu- 
tation. The  brushes  of  a  generator  are,  therefore,  moved  ahead 
in  the  direction  of  rotation  (back  in  the  case  of  a  motor) ,  so  that 
the  coil  when  short-circuited  is  cutting  the  fringe  of  lines  from 
the  pole  tip  (Fig.  220). 

If  at  the  time  represented  in  Fig.  216  there  is  an  e.m.f.  e  gen- 
erated in  the  coil  assisting  commutation,  equation  (261)  can  be 
written 


230  LECTRICAL  ENGINEERING 

This  equation  cannot  be  solved  in  general  but  it  is  possible  to  de- 
termine the  value  of  e  required  at  any  instant  to  cause  the  current 
to  vary  as  a  linear  function  of  time  from  Ic  to  —  Ic  in  the  time  T. 
On  this  assumption 

.       T  T  -  2t 
*  =  I--T- 

and  di ',        _  2I< 

dt~         T' 

Substituting  these  values  in  equation  gives 


or 

e  =  7cj^-r(l  -2^)  },  (264) 

which  gives  at  the  beginning  of  commutation  t  =  0, 

e0  =  Ic  (^  -  r)  ,  (265) 

and  at  the  end  of  commutation  t  =  T, 

eT  =  Ic  (^  +  r)  .  (266) 

This  e.m.f.  is  proportional  to  the  current  Ic  but  is  independent 
of  the  brush  resistance  rc.  The  average  value  of  this  e.m.f.  is 
called  the  commutating  e.m.f.  in  the  coil  and  is  represented  by 
Ec. 

If  the  commutating  e.m.f.  is  less  than  that  required  to  reverse 
the  current  completely  in  time  T,  commutation  is  imperfect 
and  there  is  a  tendency  to  spark,  and  if  the  e.m.f.  is  so  large  that 
the  current  is  more  than  reversed  there  is  a  tendency  to  spark  due 
to  over-commutation. 

The  contact  resistance  helps  to  prevent  sparking  when  the 
e.m.f.  generated  in  the  coil  by  rotation  is  either  too  great  or  too 
small  to  produce  perfect  commutation. 

When  commutation  is  produced  by  the  high-resistance  brush 
contact  without  the  aid  of  any  e.m.f.  generated  in  the  coil,  it  is 
called  "natural"  or  "resistance"  commutation;  when  it  is  as- 
sisted by  an  e.m.f.  generated  in  the  coil,  it  is  called  "forced"  or 
' '  voltage ' '  commutation. 

Resistance  commutation  can  never  be  perfect  unless  the  induct- 
ance of  the  coil  is  negligible,  but  at  light  loads  it  will  reverse 


DIRECT-CURRENT  MACHINERY  231 

the  current  without  injurious  sparking.  Assume  that  the  brushes 
of  a  generator  delivering  half  load  are  set  on  the  corresponding 
neutral  line  and  that  commutation  is  satisfactory.  If  the  load 
is  increased  the  increased  m.m.f.  of  the  armature  causes  the  neu- 
tral line  to  move  ahead  so  that  the  coil  short-circuited  is  cutting  a 
field  of  such  a  direction  as  to  tend  to  maintain  the  current  or  even 
to  increase  it.  The  reversal  of  the  current  is  therefore  retarded 
and  there  is  a  greater  tendency  to  spark  than  before.  If  the 
load  is  reduced  the  neutral  line  falls  behind  the  brushes  and  a 
voltage  assisting  commutation  is  generated  in  the  coil. 

Voltage  commutation  is  also  limited  in  its  application  and  as 
the  current  in  the  armature  is  increased  a  point  is  reached  (usu- 
ally about  25  per  cent,  overload)  beyond  which  sparkless  com- 
mutation is  impossible,  since  when  the  current  is  increased  a 
stronger  field  is  required  to  reverse  it,  but  the  stronger  current  in 
the  armature  increases  the  m.m.f.  of  the  armature  and  moves  the 
neutral  line  ahead  of  the  brushes  and  at  the  same  time  decreases 
the  flux.  The  brushes  have  to  be  advanced  further  and  the  de- 
magnetizing effect  is  increased.  When  the  armature  m.m.f.  is 
large  enough  to  overbalance  the  field  m.m.f.  the  flux  at  the  pole 
tip  is  wiped  out  and  voltage  commutation  is  impossible.  Moving 
the  brushes  further  ahead  only  decreases  the  flux. 

To  take  full  advantage  of  voltage  commutation  it  would  be 
necessary  to  vary  the  position  of  the  brushes  with  varying  load, 
but  this  is  not  practicable,  and  therefore  the  brushes  must  be  set 
to  give  good  commutation  at  some  intermediate  load  and  the 
resistance  of  the  brush  contact  must  be  relied  on  to  prevent  spark- 
ing above  and  below  this  point.  Modern  machines  are  designed 
to  give  good  commutation  at  all  loads  from  no  load  to  25  per  cent, 
overload  with  fixed  brush  position. 

193.  Commutating  Electromotive  Force. — The  commutating 
e.m.f.  is  the  e.m.f.  generated  in  the  short-circuited  coil  due  to 
cutting  flux  and  depends  on  a  number  of  factors. 

There  are  two  fluxes  to  be  considered:  (1)  that  produced  by  the 
field  m.m.f.;  and  (2)  that  produced  by  the  armature  m.m.f. 
These  two  fluxes  are  represented  in  Figs.  161  to  164  and  their 
resultant  in  the  interpolar  space  is  shown  in  Fig.  220.  The  e.m.f. 
generated  in  the  coil  due  to  cutting  the  field  flux  tends  to  assist 
commutation,  while  that  generated  by  cutting  the  armature  flux 
opposes  commutation.  The  resultant  of  the  two  is  called  the 
commutating  e.m.f.,  Ec.  This  e.m.f.,  therefore,  depends  on  the 


232 


ELECTRICAL  ENGINEERING 


field  excitation,  the  position  of  the  brushes,  and  the  load,  that  is, 
the  armature  current. 

When  the  brushes  are  on  the  " no-load"  neutral  there  is  no 
e.m.f.  assisting  commutation;  if  the  machine  is  carrying  load  at 
the  time  there  will  be  an  armature  flux  both  in  the  interpolar 
space  and  around  the  end  connections  and  this  flux  being  sta- 
tionary is  cut  by  the  short-circuited  coil  and  generates  in  it  an 
e.m.f.  opposing  commutation.  Ec  is  therefore  negative. 

When  the  brushes  are  moved  ahead  to  the  position  where  the 
e.m.f.  due  to  cutting  the  field  flux  is  equal  to  the  e.m.f.  due  to 
cutting  the  armature  flux,  the  commutating  e.m.f.  is  zero.  This 
position  is  called  the  load  neutral.  In  this  case  commutation  is 
carried  out  by  the  brush-contact  resistance  and  will  be  satis- 
factory only  if  the  reactance  voltage  is  very  low. 

When  the  brushes  are  moved  ahead  of  this  load  neutral  to  a 
position  where  the  commutating  e.m.f.  is  positive  and  equal  to  the 
reactance  voltage  commutation  is  perfect. 


X^C 


ER 


0   X   M  H   I   l«l^LoadO 

tttuiit  Generator 


M  *  H    llHLwuJO   M  H  K    l  i^LoadO    X   X  K 
Flat-Coinpouud  Over-Compouud  Interpole  Generator 


FIG.  221.  —  Reactance  voltage. 

In  Fig.  221  the  reactance  voltage  ER  and  the  commutating 
voltage  Ec  are  shown  plotted  on  a  load  base  for  different  types  of 
generators. 

The  reactance  voltage  is  directly  proportional  to  the  current 
and  is  represented  by  a  straight  line  passing  through  the  origin. 
For  the  shunt  generator  the  commutating  e.m.f.  varies  from  a 
maximum  at  no  load  to  zero  at  about  50  per  cent,  overload. 
The  brushes  are  set  to  give'  perfect  commutation  at  75  per  cent. 
load.  At  25  per  cent,  overload  the  armature  m.m.f.  pretty  well 
wipes  out  the  main  field  near  the  pole  tip  and  only  a  very  small 
commutating  e.m.f.  is  left.  Sparking  will  not  be  serious  so  long 
as  the  difference  between  ER  and  Ec  is  not  greater  than  the  voltage 
drop  at  one  brush  contact.  The  reactance  voltage  at  full  load 
should  be  limited  to  this  or  a  lower  value. 

For  a  flat-compounded  machine  the  conditions  are  much  better, 
the  commutating  e.m.f.  does  not  decrease  to  any  great  extent 


DIRECT-CURRENT  MACHINERY  233 

with  load  and  the  difference  between  ER  and  Ec  is  very  much  less; 
larger  values  of  reactance  voltage  may,  therefore,  be  allowed. 

For  an  over-compounded  machine  conditions  are  still  further 
improved;  Ec  increases  with  load  and  satisfactory  commutation 
may  be  obtained  with  reactance  voltages  more  than  double  the 
drop  at  the  brush  contact. 

194.  Conditions  Essential  to  Sparkless  Commutation. — It  is 
necessary  to  limit  the  armature  strength  at  full  load  to  such  a 
value  that  it  will  not  interfere  too  greatly  with  the  flux  produced 
by  the  field  winding.     For  shunt  machines  the  field  ampere- 
turns  required  to  drive  the  flux  through  the  gap  and  teeth  should 
be  =  1.2  (armature  ampere-turns  per  pole)  +  the  demagnetizing 
ampere-turns  per  pole.     This  insures  good  commutation  from 
no  load  to  25  per  cent,  overload  with  fixed  brush  position  pro- 
vided the  reactance  voltage  does  not  exceed  the  limits  specified 
below.     Lower  values  of  armature  strength  tend  to  give  better 
commutating  conditions. 

For  flat-compounded  machines  the  field  ampere-turns  for  the 
gap  and  teeth  should  be  =  1.2  (armature  ampere-turns  per  pole). 

The  reactance  voltage  must  also  be  limited  approximately  to 
the  following  values. 

In  shunt  machines  from  1  to  1.25  volts  except  with  high- 
resistance  brushes. 

In  10  per  cent,  over-compounded  generators  from  2.5  to  3.0 
volts. 

In  shunt  motors  where  any  large  variation  of  speed  by  field 
control  is  to  be  obtained  values  of  reactance  voltage  below  1 
volt  should  be  used  or  a  very  weak  armature. 

In  interpole  machines  satisfactory  commutation  may  be 
obtained  with  reactance  voltages  up  to  15  volts. 

195.  Calculation  of  the  Reactance  Voltage  for  a  Full-pitch 
Multiple  Winding. — In  Fig.  218  (a)  is  shown  one  coil,  c,  of  a  full- 
pitch  double-layer  winding,  short-circuited  by  the  brush  and  at 
(b)  is  shown  a  section  through  a  slot.     If  the  number  of  turns  per 
coil  is  n,  the  number  of  conductors  per  slot  is  2n.     Here  n  =  2. 
The  flux  of  self  and  mutual  inductance  is  shown  in  the  figure 
and  may  be  divided  into  two  parts:  (1)  that  surrounding  the  slot 
part  of  the  coil;  and  (2)  that  around  the  end  connections. 

Let  <j)S  be  the  number  of  lines  of  force  which  link  1-in.  length  of 
the  slot  part  of  the  coil  for  each  ampere  conductor  in  the  group 
of  conductors  simultaneously  short-circuited;  and  let  <j>e  be  the 


234  ELECTRICAL  ENGINEERING 

number  of  lines  of  force  which  link  1-in.  length  of  the  end  connec- 
tions for  each  ampere  conductor  in  the  corresponding  group  of 
end  connections  simultaneously  short-circuited. 

Then,  with  a  full-pitch  winding  and  a  brush  covering  only  one 
bar  the  total  inductive  flux  linking  the  coil,  c,  when  carrying  a 
current  i  amp.,  is 

6,  =  2(2m>sLc)  +  2(m>eLe)  =  2m(20sLc  +  0eLe), 
where  Lc  =    length  of  the  imbedded  part  of  one  conductor  in 
inches,  and  Le  =  length  of  one  end  connection. 

The  inductance  of  the  coil  is 

L  =  ^  10-8  =  2n*(2<t>sLc  +  <f>eLe)lQ-*  henrys 

u 

and  the  average  reactance  voltage  is 

ER=^L  (page  229). 

The  time  of  commutation  is 

width  of  brush  or  bar 


T  = 


peripheral  velocity  of  commutator 

width  of  bar 

r.p.s.  X  no.  of  bars  X  width  of  bar 
1  1 


r.p.s.  X  no.  of  bars       r.p.s.  X  no.  of  coils 


armature  conductors      r.p.m.  X  Z 
~2rT 

From  tests  made  on  a  large  number  of  machines  Hobart  found 
that  <f>,  has  a  value  of  approximately  10  and  (f>e  a  value  2. 
Therefore,  the  average  reactance  voltage  is 


7?     —  "*-c"   XN  x'fr"A^'    o^s/onr     _i_  or  ^Mn-8 
CIR  —  —     —         —  \  ZiH  \Z\jLic  "f~  Zi±je)  1 1U 

=  10   C-.^Le  10~8  X  /c^  X  r.p.m.  X  n 
lo 

= Y^ —  —  10~8  X  ampere    conductors    on .  the 

lo 

armature  X  r.p.m.  X  turns  per  coil. 

For  the  ordinary  type  of  coil  Le  may  be  taken  as  =  2LC  and 
the  average  reactance  voltage  may  be  expressed  by 

ER  =  0.8  X  10~8  X  Lc  X  ZIC  X  r.p.m.  X  n  (267) 

=  0.8  X  10~8  X  gross  length  of  core  X  ampere  conduc- 
tors on  the  armature  X  r.p.m.  X  turns  per  coil. 


DIRECT-CURRENT  MACHINERY 


235 


An  increase  in  the  brush  width  does  not  change  the  reactance 
voltage  because  although  it  increases  the  number  of  coils  short- 
circuited  at  any  instant  it  also  increases  the  time  of  commutation 
in  the  same  proportion. 

If  there  are  six  coil  sides  per  slot,  Fig.  222,  and  the  brush  covers 
three  segments  then  the  flux  surrounding  the  slot  part  of  the  coil 
will  be  three  times  as  great  as  before  since  the  m.m.f  .  is  increased 
to  three  times,  while  any  increase  in  the  width  of  the  path  across 
the  slot  will  be  counterbalanced  by  a  — 
proportional  increase  in  the  depth. 
Similarly,  the  flux  about  the  end  con- 
nections will  be  about  three  times  as 
great;  therefore,  the  inductance  is  in- 
creased three  times  but  the  time  of  com- 
mutation is  also  increased  three  times 
and  the  reactance  voltage  remains  as 
before. 

With  a  short-pitch  winding  (Fig.  219) 
the  slot  flux  is  reduced  to  about  half  the 
value  with  a  full-pitch  winding.  This  is  due  to  the  increased 
reluctance  of  the  path.  The  flux  about  the  end  connections  is 
decreased  only  slightly,  due  to  their  shorter  length. 

The  reactance  voltage  of  a  short-pitch  winding  is 

^£  10-s  x  ZIC  X  r.p.m.  X  n 


FIG.  222. 


lo 

or  again  substituting  Le  =  2LC 

ER  =  0.46  X  10-8  X  Lc  X  ZIe  X  r.p.m.  X  n.          268; 

The  ratio  -I-T  —  rrjr-was  assumed  to  be  approximately  3.5  in 

the  derivation  of  the  formulae  above.  If  very  deep  and  narrow 
slots  are  used  the  flux  <f>s  will  be  increased  and  therefore  also  the 
reactance  voltage. 

In  the  case  of  a  series  or  two-circuit  winding  there  are  p/2  coils 
short-circuited  in  series  and  the  reactance  of  this  set  of  coils  is 
p/2  times  the  reactance  of  one  coil  but  the  commutating  e.m.f  .  is 
also  increased  in  the  proportion  p/2.  Since  both  these  voltages 
are  greater  than  before  it  is  difficult  to  keep  their  difference  with- 
in the  value  which  can  be  taken  care  of  by  the  brush-contact 
resistance.  On  the  other  hand,  if  the  commutating  field  is  too 
strong,  the  reactance  of  p/2  coils  in  series  tends  to  prevent  the 


236  ELECTRICAL  ENGINEERING 

current  from  growing  to  a  large  value.  Values  of  reactance 
voltage  calculated  as  for  full-pitch  windings,  equation  (267), 
should  be  kept  about  50  per  cent,  lower  for  series  windings. 

196.  Current  Density  at  the  Brush  Contacts. — There  are  two 
currents  in  the  coil  short-circuited  by  the  brush:  (1)  the  working 
current  or  load  current;  and  (2)  the  short-circuit  current  pro- 
duced by  an  active  e.m.f.  in  the  coil. 

When  the  e.m.f.  generated  'in  the  coil  is  equal  and  opposite  to 
the  reactance  voltage  there  is  no  circulating  current  produced  in 
the  coil  and  the  load  current  reverses  at  a  uniform  rate  due  to  the 
action  of  the  contact  resistance  (page  226) ;  the  current  density 
is  then  uniform  over  the  brush  face  and  the  loss  of  energy  is  a 
minimum. 

When  the.  two  e.m.fs.  in  the  coil  do  not  balance  their  result- 
ant tends  to  cause  a  current  to  circulate  through  the  coil  and 
brush,  and  the  current  density  is  no  longer  uniform  and  the  loss 
at  the  contacts  is  increased  above  normal. 

This  circulating  current  crosses  the  brush  contact  twice  and 
the  contact  resistance  tends  to  keep  the  current  down  to  a  reason- 
able value.  When  the  unbalanced  e.m.f.  is  not  greater  than  2 
to  2.5  volts  the  current  will  not  be  large. 

Reducing  the  width  of  the  brush  may  in  some  cases  so  cut  down 
the  short-circuit  current  that  the  maximum  current  density  is 
decreased  although  the  average  is  increased. 

The  average  current  density  in  machines  without  interpoles 
is  from  35  to  40  amp.  per  square  inch  but  higher  values  are  some- 
times used  with  low-contact  resistance  brushes. 

Unequal  distribution  of  current  between  brushes  on  the  same 
arm  or  between  arms  of  the  same  polarity  causes  the  density  to 
vary  from  the  average.  This  is  particularly  true  of  machines 
with  series  windings  and  gives  rise  to  selective  commutation. 

In  the  alternating-current  series  commutator  motor  in  which 
the  circulating  currents  tend  to  become  very  large  the  contact 
resistance  is  reinforced  by  the  addition  of  high-resistance  leads 
between  the  coils  and  the  commutator  bars. 

197.  Burning  of  the  Brush  and  Commutator. — An  apparent 
electrolytic  action  takes  place  under  the  brushes  in  the  direction 
of  the  current  flow,  carbon  is  deposited  on  the  commutator  under 
the  negative  brush  and  the  brush  is  burned,  and  copper  is  de- 
posited on  the  positive  brush  and  the  commutator  is  burned. 
The  burning  is  dependent  on  the  energy  consumed  at  the  par- 


-f 


DIRECT-CURRENT  MACHINERY  237 

ticular  point  on  the  brush  contact  and  this  is  proportional  to  the 
product  of  the  current  density  at  the  point  and  the  voltage  drop 
at  the  contact. 

Severe  burning  of  the  commutator  results  in  high  mica  which 
spoils  the  contact  between  the  brush  and  the  commutator  and 
increases  the  liability  to  spark.  To  pre- 
vent the  commutator  burning  in  grooves  • 
the  brushes  should  be  staggered  in  sets  as 
shown  in  Fig.  223. 

Even  a  very  slight  deposit  of  copper  on 
the  brush  face  reduces  the  brush-contact 
resistance  very  materially.  On  the  other 
hand,  the  darkening  or  polish  on  the  commu- 
tator  surface  increases  the  contact  resistance. 

198.  Poor    Commutation    Resulting    from    Too    Many    Coil 
Sides  per  Slot. — When  a  number  of  coil  sides  are  placed  in  one 
slot,  they  are  not  all  short-circuited  at  once  and  therefore  they 
will  be  commutated  in  fields  of  different  strengths  and  the  com- 
mutating  e.m.fs.  will  be  different  since  the  intensity  of  the  field 
varies  very  rapidly  as  the  pole  tip  is  approached.     The  first 
coil  to  be  commutated  will  be  in  the  weakest  field  and  may  not 
have  its  current  completely  reversed,  while  the  last  may  be  over- 
commutated.     If  there  are  six  coil  sides  per  slot  every  third 
commutator    bar    is    liable    to    be    burned    due    to    improper 
commutation. 

It  is  therefore  not  advisable  to  reduce  the  number  of  slots 
too  far.  Small  machines  should  have  at  least  12  slots  per  pole 
and  large  machines  at  least  14  slots  per  pole. 

199.  Interpoles. — Interpoles  are  small  poles  placed  midway 
between  the  main  poles  of  either  motors  or  generators.     They  are 
magnetized  by  a  winding  connected  in  series  with  the  armature 
and  carrying  the  load  current.     Fig.  224  shows  an    interpole 
motor   or   generator.     The   brushes   are   fixed   on   the  no-load 
neutral  points.     The  interpoles  have  the  same  effect  as  moving 
the  brushes  since  they  move  the  poles  magnetically. 

The  m.m.f.  of  the  interpole  winding  must  oppose  the  m.m.f. 
of  the  armature  and  must  be  strong  enough  to  overbalance  it 
and  produce  a  field  under  the  interpole  of  the  proper  intensity  to 
reverse  the  current  in  the  short-circuited  coil.  Since  the  inter- 
pole winding  is  in  series  with  the  armature  the  commutating  field 
increases  with  load  and  satisfactory  commutation  up  to  and 


238  ELECTRICAL  ENGINEERING 

beyond  the  overload  limits  of  output  set  by  armature  heating 
can  be  obtained. 

The  interpole  provides  a  much  more  definite  means  of  con- 
trolling the  commutating  e.m.f.  than  is  the  case  where  reliance 
is  placed  on  shifting  the  brushes.  As  the  output,  the  speed  or  the 
voltage  of  direct-cunent  machines  is  increased  the  reactance 
voltage  is  increased  and  a  limit  is  reached  beyond  which  satis- 
factory commutation  can  no  longer  be  obtained  economically  by 
shifting  the  brushes  and  interpoles  must  be  supplied.  They  are 
required  for  all  large  reversible  motors,  where  the  brushes  must 
be  fixed  on  the  no-load  neutral  and  where  therefore  no  commutat- 
ing e.m.f.  can  be  obtained  without  them.  They  are  also  required 


FIG.  224. — Interpole  generator  or  motor. 

for  adjustable-speed  shunt  motors  where  a  large  speed  range  is 
to  be  obtained  by  field  control.  With  interpoles  a  speed  range  of 
3  to  1  or  even  4  or  5  to  1  is  possible. 

Interpoles  are  not  so  necessary  on  generators  as  on  motors  but 
they  are  useful  where  very  wide  variations  of  load,  are  met  with 
and  especially  if  the  variations  are  rapid.  The  interpole  m.m.f. 
follows  exactly  the  fluctuations  of  the  armature  m.m.f.  but  if  the 
changes  of  load  are  very  rapid,  it  is  difficult  for  the  interpole  flux 
to  build  up  rapidly  enough  to  give  perfect  commutation,  since 
this  flux  has  to  pass  through  the  solid  yoke  and  eddy  currents 
are  induced  which  tend  to  retard  it. 

If  the  interpole  is  provided  with  a  shunt,  it  is  necessary  to 
have  the  resistance  of  the  shunt  so  adjusted  that  the  interpole 
winding  will  receive  its  proper  proportion  of  the  current,  but  in 
addition  the  shunt  must  be  designed  with  an  inductance  propor- 
tional to  the  inductance  of  the  interpole  winding  to  insure  the 
proper  division  of  the  current  when  the  load  is  fluctuating 
rapidly. 


DIRECT-CURRENT  MACHINERY  239 

Interpoles  must  be  designed  with  very  low  flux  densities  at 
normal  load,  so  that,  under  heavy  overloads,  they  will  not  be- 
come saturated  and  will  be  able  to  carry  the  flux  required  to 
produce  the  increased  commutating  e.m.f.  Interpoles  should 
be  provided  on  all  machines  where  the  reactance  voltage  is  over 
3.5  or  4  volts  and  with  them  satisfactory  commutation  can  be 
obtained  with  reactance  voltages  up  to  15  volts.  The  interpole 
shunt  must,  however,  be  very  carefully  adjusted  or  the  difference 
between  the  commutating  voltage  and  the  reactance  voltage 
may  be  so  large  that  serious  circulating  currents  will  be 
produced. 

When  interpoles  are  provided  the  ratio  of  the  field  ampere- 
turns  per  pole  for  the  gap  and  teeth  to  the  armature  ampere- 
turns  per  pole  may  be  as  low  as  0.8  and  it  is  no  longer  necessary 
to  limit  the  armature  strength  to  such  a  low  value  as  that  given 
for  non-interpole  machines  on  page  261. 

The  interpole  m.m.f.  wipes  out  the  armature  flux  in  the  com- 
mutating zone  but  does  not  prevent  the  distortion  of  the  flux 
under  the  poles  (Fig.  224).  This  distortion  is  more  serious  be- 
cause of  the  increased  armature  strength  and  is  liable  to  produce 
flashing  due  to  the  generation  of  excessive  voltages  between 
adjacent  commutator  bars  (see  Art.  200). 

200.  Flashing. — When  an  arc  passes  over  the  surface  of  the 
commutator  from  a  positive  to  a  negative  brush,  the  machine  is 
said  to  "flashover."  A  severe  flash  is  equivalent  to  a  dead 
short-circuit  and  may  have  serious  effects. 

Flashing  is  usually  due  to  a  combination  of  a  high  voltage 
between  adjacent  commutator  bars  with  a  partial  short-circuit 
between  bars  due  to  the  collection  of  carbon  or  graphite  dust  on 
the  mica  insulation. 

The  causes  of  flashing  are  distinct  from  the  causes  producing 
sparking  at  the  brushes  but  a  flash  may  sometimes  be  started 
by  a  spark  originating  under  the  brush  due  to  poor  commutation. 

Fig.  225  shows  the  flux  distribution  in  the  air  gap  at  no  load 
and  at  full  load.  At  no  load  the  flux  density  under  the  pole 
face  is  uniform  and  the  voltage  generated  between  bars  is  com- 
paratively low;  under  load  the  armature  m.m.f.  causes  the  flux 
density  to  increase  at  one  pole  tip  ti  and  to  decrease  at  the  other 
tip  t.  The  voltage  between  bars  at  t\  will  be  very  much  increased 
and,  if  the  commutator  is  dirty,  it  may  cause  a  flash  locally  be- 
tween bars  and  this  flash  may  be  carried  around  in  the  direction 


240  ELECTRICAL  ENGINEERING 

of  rotation  and  result  in  a  complete  short-circuit  between  brushes 
either  momentary  or  permanent  depending  on  conditions. 

When  the  load  comes  on  a  motor  very  suddenly  the  armature 
m.m.f.  causes  a  part  of  the  flux  to  swing  quickly  across  from 
t  to  ti  and,  as  this  movement  is  opposite  to  the  direction  of  rota- 
tion, the  flux  is  cut  more  rapidly  than  under  ordinary  conditions 
and  the  tendency  to  flash  is  increased.  In  the  case  of  a  generator 
the  swing  of  flux  is  in  the  direction  of  rotation  and  the  voltages 
are  not  so  high. 

If  the  load  is  suddenly  removed  from  a  generator  the  swing  of 
flux  is  opposite  to  the  direction  of  rotation  and  the  voltage  be- 
tween bars  may  be  very  high. 

Full  Load  Field  Form 
No-Load  Field  Form 


FIG.  225. 

When  the  magnetic  circuit  is  highly  saturated  and  the  machine 
has  what  is  called  a  stiff  field,  the  flux  peak  at  ti  is  not  so  high  and 
there  is  less  danger  of  flashing. 

A  motor  operating  at  high  speed  due  to  field  weakening  is 
peculiarly  liable- to  flash;  the  reduced  saturation  allows  the  arma- 
ture m.m.f.  to  produce  a  high  flux  peak  while  the  increased  speed 
counterbalances  the  decreased  main  flux;  and  at  the  same  time 
commutating  conditions  are  bad. 

Interpole  machines  are  more  likely  to  flash  than  those  without 
interpoles  because  of  their  increased  armature  strength.  In 
non-interpole  machines  the  field  m.m.f.  per  pole  is  from  30  to 
40  per  cent,  greater  than  the  armature  m.m.f.  per  pole  while  in 
interpole  machines  the  field  m.m.f.  per  pole  is  approximately 
equal  to  the  armature  m.m.f.  and  may  even  be  less.  Commuta- 
tion is  taken  care  of  by  the  interpole  flux  but  the  armature  m.m.f. 
which  is  all  cross-magnetizing,  since  the  brushes  are  on  the  no- 
load  neutral,  produces  a  large  distortion  of  the  flux  under  the 
poles  due  to  the  relatively  great  armature  strength. 


DIRECT-CURRENT  MACHINERY  241 

If  load  is  suddenly  applied  to  an  interpole  motor  the  flux  swings 
rapidly  across  the  pole  and  piles  up  at  ti  and  induces  a  high  volt- 
age between  bars ;  at  the  same  time  sparking  at  the  brushes  may 
occur  due  to  the  fact  that  the  interpole  commutating  flux  cannot 
build  up  as  fast  as  the  armature  current  since  the  interpole  field 
winding  has  a  large  inductance  and  the  solid  yoke  opposes  the 
required  change  of  flux  by  the  production  of  eddy  currents.  A 
flash  may  therefore  result.  Further,  if  the  interpole  shunt  is  not 
properly  designed  it  may  happen  that  the  rapidly  changing  cur- 
rent will  follow  the  path  of  low  inductance  through  the  shunt 
rather  than  the  proper  path  through  the  winding.  Interpoles 
should  be  designed  with  inductive  shunts  to  take  care  of  such 
currents. 

In  the  case  of  an  interpole  generator  flashing  is  more  severe 
when  load  is  removed  suddenly  as  the  swing  of  flux  is  then  in  the 
direction  to  produce  the  greatest  voltage  between  bars;  at  the 
same  time  the  interpole  commutating  field  cannot  disappear 
immediately  and  sparking  occurs  at  the  brushes  due  to  over- 
commutation  of  the  current. 


Fall  Load  Field  Form 
No-Load  Field  Form 


FIG.  226. — Compensating  winding. 

Where  extreme  variations  of  load  are  liable  to  occur  with  great 
rapidity  it  is  advisable  to  supply  the  machines  with  compensating 
windings  placed  in  slots  in  the  pole  faces,  Fig.  226,  to  counteract 
the  armature  m.m.f.  over  the  whole  armature  surface  and  so 
prevent  any  distortion  of  the  flux. 

The  maximum  value  of  volts  per  bar  under  normal  operating 
conditions  should  not  exceed  30  volts  for  moderate  capacities 
or  28  volts  for  large  capacities.  The  average  volts  per  bar  will 
be  about  half  of  the  maximum  value  and  should  not  exceed  about 
15  volts.  For  small  machines  operating  under  steady  load  higher 
values  may  be  used  but  the  smaller  values  are  preferable  if  they 

16 


242 


ELECTRICAL  ENGINEERING 


can  be  obtained  without  too  great  a  sacrifice,  especially  when  the 
machine  may  experience  sudden  changes  of  load. 

201.  Compensating  Windings. — In  cases   where   the  average 
voltage  between  bars  is  already  so  high  that  it  is  necessary  to 
limit  the  field  distortion  to  prevent  flashing,  compensating  wind- 
ings should  be  provided.     They  are  windings  placed  in  slots  in 
the  pole  faces,  which  exert  a  m.m.f.  equal  and  opposite  to  the 
armature  m.m.f.  and  so  prevent  any  distortion  of  the  field.     The 
compensating  winding  carries  the  load  current  and  may  there- 
fore be  designed  with  a  smaller  number  of  turns  than  the  arma- 
ture.    Interpoles  are  required  in  addition  to  provide  the  corn- 
mutating  e.m.f . ;  but  since  the  armature  m.m.f.  is  already  neutral- 
ized by  the  compensating  winding,  the  m.m.f.  of  the  interpole 
winding  is  only  that  required  to  produce  the  flux  necessary  for 
commutation,  through  the  magnetic  circuit  of  the  interpole. 
Fig.  226  shows  a  machine  provided  with  interpoles  and  a  com- 
pensating winding. 

Generators  with  compensating  windings  require  only ,  weak 
series  fields  to  maintain  constant  terminal  voltage  since  the 
armature  reaction  is  neutralized.  The  resistance  drop  is  in- 
creased to  a  slight  extent. 

202.  Parallel  Operation. — In  power  houses  in  which  the  load 
varies  at  different  hours  of  the  day  a  number  of  generators  are 

usually  installed.  When  the 
load  is  light  one  generator  is 
operated  and  supplies  the 
demand  and  when  the  load 
increases  a  second  machine  is 
started  up  and  connected  in 
parallel  with  the  first  and  its 
excitation  is  adjusted  until  it 
takes  its  proper  share  of  the 
load.  Fig.  227  shows  two 
shunt  generators  (1)  supplying 

power  and  (2)  ready  to  be  connected  in  parallel  with  it.  Before 
closing  the  switches  Si  and  Sz  which  connect  the  second  machine 
to  the  load  it  is  necessary  that  its  polarity  be  correct  and  that  its 
terminal  voltage  be  the  same  or  a  little  higher  than  that  of  (1). 
If  the  field  rheostat  of  (2)  is  so  adjusted  that  the  voltage  of  (2) 
is  the  same  as  the  voltage  of  (1)  and  switch  Si  is  closed,  then,  if 
there  is  no  voltage  across  $2,  it  may  be  closed.  But  if  the  voltage 


8. 


IjmjOQOO  J  UlflJLfiJLfiJL 

.FiG.  227. — Parallel  operation  of  shunt 
generators. 


DIRECT-CURRENT  MACHINERY 


243 


across  S%  is  found  to  be  about  double  the  terminal  voltage,  the 
polarity  of  (2)  must  be  reversed  before  closing  switch  $2-  After 
closing  £2  the  field  rheostat  of  (2)  must  be  adjusted  until  (2)  takes 
its  proper  share  of  the  load. 

If  the  voltage  of  (2)  is  the  same  as  the  voltage  of  (1)  when 
the  switch  is  closed,  (2)  will  not  take  any  load  but  will  run  idle. 
If  the  voltage  of  (2)  is  less  than  the  voltage  of  (1),  machine  (2) 
will  run  as  a  motor  driving  its  prime  mover  and  will  draw  power 
from  (1).  If,  however,  the  terminal  voltage  of  (2)  is  higher  than 
that  of  (1),  machine  (2)  will  supply  part  of  the  load  and  will 
relieve  (1)  until  the  voltages  of  the  two  become  the  same.  Fig. 
228  represents  the  voltage  characteristics  of  the  two  machines 
plotted  on  the  same  base.  If  the  terminal  voltage  is  E,  (1)  sup- 
plies a  current  Ji  and  (2)  a  current  72  and  the  total  current 


••d 

^ 

==: 
"—  -. 

(1) 

^ 

'  !• 

(2) 
C 

—  -^.^ 

—  •      — 

(3) 

•  — 

--.  ^^ 

To- 

—  » 

~"-< 

^" 

^^ 

T 

j    j 

i+T 

Amperes 
(1)  Generator  (1) 
(2)   Generator  (2') 
(3)   Station 

FIG.  228. 


Amperes 

(1)  Generator  (1) 

(2)  Generator  (2) 

(3)  Station 

Fia.  229. 


supplied  by  the  station  is  /  =  7i  +  /2.  The  machine  with  the 
flatter  characteristic  will  supply  the  greater  amount  of  power. 
If  the  two  machines  are  rated  at  the  same  current  output,  (2) 
can  be  made  to  take  its  share  of  the  load  by  cutting  out  resist- 
ance from  its  field  rheostat  and  so  raising  its  voltage  character- 
istic and  inserting  resistance  in  the  field  circuit  of  (1)  and  lowering 
its  characteristic  as  shown  in  Fig.  229. 

Shunt  generators  will  operate  in  parallel  and  divide  up  the 
load  in  proportion  to  their  capacities  if  their  voltage  character- 
istics are  similar,  that  is,  if  their  terminal  voltage  falls  from  no 
load  to  full  load  by  the  same  amount  and  in  the  same  manner. 
If  the  characteristics  are  different  a  proper  division  of  load  can  be 
obtained  by  regulating  the  field  rheostats. 

203.  Parallel  Operation  of  Compound  Generators. — Fig.  230 
shows  two  compound-wound  generator's  connected  in  parallel  and 


244 


ELECTRICAL  ENGINEERING 


supplying  power  to  a  load  circuit.  Their  voltage  characteristics 
are  shown  in  Fig.  231.  Assume  that  the  prime  mover  of  (1)  runs 
for  an  instant  at  a  slightly  increased  speed;  the  voltage  of  (1) 
rises  and  it  takes  more  than  its  share  of  the  load;  the  voltage  of 
(2)  falls  because  its  load  is  decreased  and  its  series  excitation  is 
decreased.  Machine  (1)  therefore  takes  more  of  the  total  load 
and  its  voltage  rises  higher  until  it  supplies  all  the  load  and  in 


FIG.  230. 


addition  drives  (2)  as  a  motor.  Since  the  current  in  (2)  is  re- 
versed the  m.m.f .  of  its  series  winding  is  also  reversed  and  it  runs 
as  a  differential  motor  driving  its  prime  mover  at  a  high  speed 
until  the  load  on  (1)  becomes  so  great  that  the  protective  appa- 
ratus opens  the  circuit  and  shuts  down  the  system. 

To  get  over  this  difficulty  the  equalizer  connection  ee,  Fig.  232, 
is  used.  It  is  a  conductor  of  low  resistance  connecting  the  series 
windings  of  the  two  machines  in  multiple.  Now  if  the  prime  mover 

of  (1)  runs  above  normal  speed 
the  voltage  of  (1)  rises  and  it 
takes  an  increased  load.  The 
increase  of  current  does  not 
all  go  through  the  series  wind- 
ing of  (1)  but  divides  between 
the  windings  of  (1)  and  (2)  in 
inverse  proportion  to  their  re- 
sistances and  so  prevents  any 
decrease  of  the  voltage  of  (2) . 
Thus  with  an  equalizer  con- 
nection (2)  will  still  hold  its 
load.  The  resistances  of  the  series  windings  must  be  adjusted  so 
that  the  load  current  will  divide  between  them  in  such  propor- 
tion that  each  machine  will  supply  its  proper  share  of  the  load. 
Before  connecting  machine  (2)  in  parallel  with  (1)  which  is 
delivering  power,  first  close  switches  >S2  and  /S3,  Fig.  232,  and 


Si 

<S 

S2 

«i 

<s> 

JULfiJL&JUL 

e 

m 

^3 
Equalizer  Connection 

.  .       £ 

€> 

j.ooojm. 

w 

FIG.  232. — Parallel  operation  of  com- 
pound  generators. 


DIRECT-CURRENT  MACHINERY  245 

adjust  the  shunt  field  of  (2)  until  its  terminal  voltage  is  the  same 
as  that  of  (1).  The  excitation  of  (2)  is  now  provided  partly  by 
its  shunt  field  and  partly  by  its  series  field  carrying  part  of  the 
load  current.  After  checking  the  polarity  to  see  that  it  is  correct 
close  switch  Si  and  adjust  the  shunt  field  of  (2)  until  the  machines 
divide  the  load  in  proportion  to  their  capacities. 

From  the  above  discussion  it  is  seen  that  two  compound- 
wound  generators  connected  in  parallel  form  an  unstable  system 
unless  an  equalizer  connection  is  placed  between  their  series 
windings. 

204.  Storage  Batteries. — A  storage  battery  is  an  apparatus 
in  which  electrical  energy  can  be  stored  to  be  used  at  some  later 
time. 

Batteries  are  made  up  of  a  number  of  cells  connected  in  series 
multiple  according  to  the  voltage  and  current  required. 

Each  cell  is  composed  of  two  plates  or  electrodes  of  suitable 
materials  immersed  in  an  electrolyte.  The  most  commonly  used 
storage  battery  has  a  positive  plate  of  lead  peroxide  Pb02  and  a 
negative  plate  of  sponge  lead  Pb  immersed  in  dilute  sulphuric  acid 
H2SO4. 

When  the  battery  is  discharging  the  electrolyte  combines  with 
the  active  materials  of  the  electrodes  and  when  it  is  being  charged 
the  electrodes  are  reduced  to  their  original  condition  and  the 
materials  taken  from  the  electrolyte  are  returned  to  it. 

The  main  chemical  changes  taking  place  are  represented  by  the 
following  formula: 

charge 


PbO2  +  Pb  +  2H2SO4  =  2PbS04  +  2H2O.          (269) 


discharge 

Capacity.- — The  unit  of  capacity  of  a  storage  cell  is  the  ampere- 
hour  and  it  is  generally  based  on  the  8-hr,  discharge  rate.  An 
800-amp.-hr.  battery  will  give  a  continuous  discharge  of  100 
amp.  for  8  hr.  If,  however,  the  rate  of  discharge  is  increased  the 
ampere-hour  capacity  of  the  battery  decreases.  At  a  6-hr,  dis- 
charge rate  the  capacity  is  only  about  95  per  cent.,  at  a  4-hr, 
discharge  rate  it  is^about  80  per  cent,  and  at  a  1-hr,  rate  it  is  only 
50  per  cent,  of  its  8-hr,  rating.  Thus  the  battery  mentioned  above 
would  give  a  continuous  discharge  of  400  amp.  for  only  1  hr. 


246 


ELECTRICAL  ENGINEERING 


The  capacity  of  a  cell  is  proportional  to  the  area  of  the  plates 
exposed  to  the  electrolyte  and  for  an  8-hr,  discharge  rate  a 
current  density  of  from  40  to  60  amp.  per  square  foot  of  positive 
plate  is  common  practice. 

Voltage. — The  voltage  of  a  cell  depends  on  the  character  of  the 
electrodes,  the  density  of  the  electrolyte  and  the  condition  of  the 
cell  but  is  independent  of  the  size.  The  variation  of  the  terminal 
voltage  of  a  cell  during  charge  and  discharge  is  shown  by  the 
curves  in  Fig.  233.  On  charge  the  voltage  begins  about  2  volts 
and  rises  to  2.5  volts  or  a  little  above.  When  the  charging  cir- 
cuit is  opened  the  voltage  falls  to  2.1  volts  and  during  discharge 
falls  off  gradually  to  about  1.9  volts.  Beyond  this  point  the  fall 
of  voltage  is  very  rapid  and  discharge  should  not  be  continued 
after  the  voltage  has  fallen  to  1.7  volts. 


-==-  Battery 

EM:     Variable  Load 


FIG.  233. — Voltage  characteristics  of  a      FIG.  234. — Battery  with  a  shunt- 
storage  cell.  generator. 

The  required  battery  voltage  is  obtained  by  connecting  a  num- 
ber of  cells  in  series  and  the  required  current  is  obtained  by  con- 
necting a  number  of  plates  or  cells  in  multiple. 

205.  Applications. — Batteries  are  installed  in  direct-current 
power  stations  to  store  energy  during  periods  of  light  load  and  to 
deliver  energy  in  parallel  with  the  generators  during  periods  of 
heavy  load.  When  the  load  is  light  the  generators  charge  the 
battery  and  when  the  load  is  heavy  the  charge  is  given  up  and  so 
the  load  on  the  generators  is  maintained  nearly  constant  and  they 
can  be  operated  at  maximum  efficiency.  The  result  is  that  the 
voltage  regulation  of  the  system  is  improved. 

Batteries  are  also  installed  in  electric-railway  substations  to 
prevent  large  variations  of  the  load  on  the  feeders  supplying 
them  and  so  regulate  the  substation  voltage. 

A  third  very  important  application  of  storage  batteries  is  in  la- 


DIRECT-CURRENT  MACHINERY 


247 


ternating-current  power  stations  where  they  provide  an  auxiliary 
supply  of  direct  current  in  case  of  a  breakdown  of  the  exciters  and 
may  thus  prevent  a  shutdown  of  the  whole  system. 

Fig.  234  shows  a  battery  connected  across  the  terminals  of  a 
shunt  generator.  At  normal  load  the  battery  voltage  and  the 
generator  voltage  are  equal  and  the  battery  floats  on  the  line 
neither  giving  nor  receiving  power.  If,  however,  the  load  in- 
creases the  generator  voltage  falls  and  the  battery  discharges  and 
supplies  part  of  the  extra  load  and  so  relieves  the  generator  and 
prevents  any  large  drop  in  its  terminal  voltage.  The  battery  in 
this  way  takes  care  both  of  sudden  overloads  and  continuous 
overloads.  During  periods  of  light  load  the  generator  voltage 
rises  above  the  battery  voltage  and  the  battery  charges. 

206.  Boosters. — Boosters  are  direct-current  generators  con- 
nected in  series  with  the  line  to  raise  or  lower  the  voltage.  They 
are  low-voltage  machines  of  large  current  capacity  and  are 
usually  driven  at  constant  speed  by  shunt  motors. 


-=-Battery 


Load 

> 

* 


Booster 
FIG.  235.— Shunt  booster. 


Load, 


FIG.  236. — Compound  booster. 


Boosters  are  used  to  compensate  for  line  drop  in  distributing 
systems  by  adding  an  equal  voltage  to  the  circuit  and  they  are 
also  used  very  extensively  to  regulate  the  charge  and  discharge 
of  storage  batteries  in  parallel  with  the  generators  in  constant- 
voltage  systems.  They  may  be  either  shunt,  series,  or  compound- 
wound. 

The  shunt  booster,  Fig.  235,  is  an  ordinary  generator  with  its 
field  connected  across  the  station  busbars  and  its  armature  in 
series  with  the  generator  armature.  Its  function  is  to  raise  the 
voltage  impressed  on  the  battery  in  order  to  send  current  into 
it  to  charge  it.  The  booster  voltage  is  controlled  by  a  field 
rheostat. 

Compound  boosters  are  automatic  in  their  action  and  are 
divided  into  two  classes,  non-reversible  and  reversible,  depending 
on  the  relative  strengths  of  their  shunt  and  series  windings. 


248 


ELECTRICAL  ENGINEERING 


Fig.  236  shows  a  non-reversible  automatic  booster.  The 
shunt  field  /  is  connected  across  the  station  busbars;  the  series 
field  s  carries  the  load  current  of  the  generator  and  it  opposes  the 
shunt  field  but  has  at  all  times  a  smaller  m.m.f.  and  thus  the 
booster  voltage  is  always  in  the  direction  of  the  generator 
voltage. 

When  the  load  current  increases,  the  increase  of  current 
through  the  series  field  decreases  the  booster  voltage  and  allows 
the  battery  to  discharge;  when  the  load  decreases,  the  booster 
voltage  rises  and  causes  the.  battery  to  charge.  The  current 
from  the  generator  remains  practically  constant  regardless  of  the 
fluctuations  of  the  load  on  the  system. 

The  reversible  booster  is  similar  in  construction  to  the  non- 
reversible  booster  but  has  a  stronger  series  field.  At  normal  load 
the  shunt  and  series  fields  are  of  equal  strength  and  the  booster 
voltage  is  zero.  The  battery  then  floats  on  the  line  and  neither 
charges  nor  discharges.  An  increase  of  load  above  normal 
increases  the  strength  of  s  and  overpowers  /  and  so  discharges 
the  battery.  When  the  load  decreases  s  becomes  weaker  than 
/  and  the  booster  causes  the  battery  to  charge. 


Booster 


FIG.  237. — Series  booster. 


Insulated  Negative  Feeder 
Negative  Booster 

FIG.  238. — Negative  booster. 


Thus  by  installing  an  automatic  booster  the  battery  is  made 
more  sensitive  to  variations  of  load  and  a  better  regulation  of  the 
generator  load  and  voltage  is  obtained. 

If  the  load  is  a  combined  lighting  and  power  load  the  lighting 
circuit  can  be  supplied  at  constant  voltage  from  the  generator 
terminals  and  the  power  circuit  connected  outside  the  battery. 
Sudden  variations  of  load  will  be  cared  for  by  the  battery  and  the 
voltage  on  the  lighting  circuit  will  not  be  affected. 

Series  boosters  are  used  in  electric-railway  engineering  and  in 
general  power  distribution  to  raise  the  voltage  on  certain  sections 
of  the  line  as  in  the  case  of  a  long  feeder  supplying  power  to  an 
outlying  section,  as  shown  in  Fig.  237.  The  booster  field  is 


DIRECT-CURRENT  MACHINERY  249 

connected  in  series  with  the  line  and  its  voltage  increases  with 
the  load  and  so  neutralizes  the  line  drop. 

In  cases  where  insulated  negative  feeders  are  connected  to  the 
track  return  at  various  points  to  carry  the  current  back  to  the 
substations  and  so  reduce  the  danger  of  electrolysis,  a  booster 
connected  in  series  with  the  negative  feeder  can  be  made  to 
reduce  its  effective  resistance  to  a  very  low  value  and  so  increase 
its  capacity.  The  use  of  such  negative  boosters,  Fig.  238,  is 
very  common  and  results  in  a  large  saving  in  conductor  material 
in  the  return  feeders. 

207.  Balancers. — A  direct-current  compensator  or  balancer 
comprises  two  or  more  similar  direct-current  machines  directly 
coupled  to  each  other  and  connected  in  series  across  the  outer 
conductors  of  a  multiple-wire  system  of  distribution  for  the  pur- 
pose of  maintaining  the  potentials  of  the  intermediate  wires. 
They  may  be  either  shunt-wound  or  compound-wound. 


FIG.  239. — Balancers. 

The  generator  G,  in  Fig.  239 (a),  develops  a  voltage  of  230 
volts  between  the  outer  wires  and  the  function  of  the  balancer 
set  BiBz  is  to  maintain  the  potential  of  the  neutral  wire  midway 
between  the  outers.  When  the  loads  on  the  two  sides  are  bal- 
anced, no  current  flows  in  the  neutral  wire  and  the  two  machines 
run  light  as  motors;  when  the  load  on  one  side  is  heavier  than  on 
the  other,  as  shown,  the  voltage  across  BI  is  lower  than  that 
across  B2  and  the  neutral  point  is  shifted;  the  machine  B2  runs 
as  a  motor  driving  B\  as  a  generator  and  partially  restores  the 
balance.  • 

The  connection  shown  in  Fig.  239(6)  gives  better  results. 
The  field  of  the  motor  B2  is  excited  from  the  heavily  loaded  side 
and  it  therefore  tends  to  run  at  an  increased  speed,  while  the  field 
of  BI  is  excited  from  the  lightly  loaded  side  and  its  voltage  there- 
fore rises  due  to  increase  in  speed  and  increase  in  excitation. 
The  balance  cannot,  however,  be  perfect  since  a  slight  inequality 
of  voltages  is  necessary  to  make  the  balancer  act. 


250 


ELECTRICAL  ENGINEERING 


Where  perfect  balance  is  required  the  machines  BI  and  B2 
must  be  compound-wound  and  connected  as  shown  in  Fig.  239(c). 
The  current  in  the  neutral  wire  flows  through  the  series  windings 
of  the  two  machines,  strengthening  the  field  of  BI  and  raising  the 
voltage  and  reducing  the  field  of  B2  and  increasing  the  speed  of 
the  set.  The  automatic  balancing  action  in  this  case  depends 
on  an  unbalance  of  current  in  the  two  outers  and  not  on  an 
unbalance  of  voltage  between  the  outers  and  neutral  as  in  (a) 
and  (6). 

208.  Rosenberg  Generator  for  Train  Lighting. — For  train 
lighting  it  is  necessary  to  have  a  generator  which  will  give 
approximately  constant  current  or  voltage  independent  of  the 
speed.  The  Rosenberg  generator  shown  diagrammatically  in 
Fig.  240  gives  a  constant  current  at  all  speeds  above  a  certain 
minimum. 

A  A  is  a  battery  which  supplies  the  lighting  system  when  the 
train  is  at  rest  or  running  at  slow  speeds.  The  field  winding  // 

is  excited  by  the  battery  and 
its  m.m.f.  Mf  remains  con- 
stant. The  main  brushes 
which  supply  the  load  are 
placed  under  the  centers  of 
the  poles  and  are  connected 
to  the  battery  terminals  and 
to  the  load,  brush  B%  directly 
and  brush  BI  through  the 
aluminum  cell  C,  which  has 
the  property  of  allowing  cur- 
rent to  flow  from  the  genera- 
tor to  the  battery  or  load  but 
offers  a  high  resistance  to  the 
flow  of  current  in  the  opposite 

direction.  The  auxiliary  brushes  bb  are  placed  on  the  commuta- 
tor at  right  angles  to  the  main  brushes  and  are  short-circuited. 
As  the  armature  rotates  its  conductors  cut  the  flux  produced 
by  the  field  m.m.f.  Mf  and  a  voltage  is  generated  between  the 
brushes  bb  but  none  between  the  main  brushes  BiB2.  A  short- 
circuit  current  flows  through  the  armature  winding  and  the 
brushes  bb  and  exerts  a  m.m.f.  Mb  at  right  angles  to  the  field 
m.m.f.  This  m.m.f.  produces  a  comparatively  large  flux. 
06  through  the  armature  and  pole  faces.  This  cross-flux  is 


FIG.  240. — Rosenberg  train-lighting 
generator. 


DIRECT-CURRENT  MACHINERY 


251 


cut  by  the  armature  conductors  and  a  voltage  is  generated  be- 
tween the  brushes  BiB2  and  a  current  /  flows  to  the  load.  Cur- 
rent /  flowing  in  the  armature  conductors  exerts  a  m.m.f.  MB 
opposing  the  field  m.m.f.  M/  and  reducing  it  to  a  comparatively 
small  value. 

Above  a  certain  speed,  which  depends  on  the  field  excitation, 
the  current  remains  approximately  constant  independent  of  the 
speed. 

The  load  current  is  limited  by  the  fact  that  the  m.m.f.  MB 
must  always  be  less  than  Mf  in  order  that  the  m.m.f.  Mb  and  the 
cross-flux  06  may  exist. 

If  the  direction  of  rotation  changes,  the  direction  of  the  e.m.f. 
between  bb  changes,  the  directions  of  Mb  and  <£&  change  and 
therefore  the  direction  of  the  voltage  from  B%  to  BI  remains  as 
before. 

Notches  are  cut  in  the  pole  faces  above  the  brushes  to  prevent 
large  voltages  being  generated  in  the  coils  which  are  being 
commutated. 

The  current  output  may  be  varied  by  varying  the  excitation. 


.Field  Winding 


(  Ct)  Kadial  Type  (&)    Axial  Type 

FIG.  241. — Homopolar  generators. 


209.  Homopolar  Generators. — Homopolar  or  acyclic  genera- 
tors are  designed  with  the  armature  conductors  revolving  in  a 
unidirectional  field  and  thus  the  e.m.fs.  generated  in  them  do  not 
alternate  during  the  revolution.  Two  types  have  been  de- 
veloped, the  radial  and  the  axial,  so-called  from  the  direction  of 
flow  of  the  armature  currents. 

In  the  radial  type,  Fig.  241  (a),  the  armature  consists  of  one 
or  more  discs  rotating  in  the  magnetic  field  and  an  e.m.f.  is 
generated  between  the  center  and  the  edge  of  the  disc.  Current 
is  collected  by  brushes  rubbing  on  the  external  periphery  of  the 
discs. 


252  ELECTRICAL  ENGINEERING 

The  e.m.fs.  generated  are  very  low  and  a  number  of  discs,  each 
supplied  with  brushes,  must  be  connected  in  series  making  a  very 
complex  and  expensive  construction. 

In  the  axial  type,  Fig.  241(6),  the  discs  are  replaced  by  one  or 
more  copper  cylinders  with  brushes  at  each  end.  The  direction 
of  the  armature  current  is  along  the  axis  of  the  machine.  Here 
again  the  e.m.f.  is  low  and  it  is  difficult  to  arrange  enough  cylin- 
ders in  series  to  give  the  required  e.m.f. 

Such  machines  can  be  used  where  very  large  currents  at  low 
voltage  are  required  as  in  electrolytic  processes. 

210.  Limits  of  Output  of  Electric  Machines.  -The  factors 
which  limit  the  output  of  electric  machines  are: 

1.  Regulation. 

2.  Efficiency. 

3.  Heating. 

4.  Commutation. 

1.  In   motors   the   regulation   is   a   speed   regulation.     With 
increase  of  load  the  speed  falls  off  and  the  increased  torque  is 
obtained  at  a  decreased  speed.     In  constant-speed  work  the 
shunt  motor  is  used  but  if  it  is  overloaded  its  speed  falls  to  a 
value  too  low  for  satisfactory  operation. 

In  generators  the  regulation  is  a  voltage  regulation.  As  the 
load  is  increased  the  voltage  falls  off  and  a  point  is  finally 
reached  where  the  voltage  is  so  low  that  the  power  supplied  is 
unsatisfactory. 

2.  The  efficiency  of  a  machine  increases  with  increasing  load 
to  the  point  where  the  variable  copper  losses  are  equal  to  the  con- 
stant losses.     Above  this  point  the  efficiency  decreases  due  to  the 
rapid  increase  of  the  variable  losses. 

With  properly  designed  machines  the  output  is  limited  by  either 
heating  or  commutation  beiore  the  regulation  or  efficiency  becomes 
too  poor. 

3.  All  the  losses  of  power  in  a  machine  are  converted  into  heat 
and  raise  the  temperature  of  the  various  parts  until  the  point  is 
reached  where  the  rate  at  which  heat  is  being  radiated  or  carried 
off  by  the  ventilating  apparatus  is  equal  to  the  rate  at  which  heat 
is  being  generated.     The  temperature  will  then  remain  constant. 
When  a  machine  is  overloaded  its  losses  increase  and  consequently 
its  temperature  rises  above  normal. 


DIRECT-CURRENT  MACHINERY  253 

If  a  machine  operates  at  a  high  temperature  for  any 
length  of  time  permanent  injury  to  the  insulating  materials 
will  result. 

4.  Sparking  will  occur  in  a  machine  when  the  field  cut  by  the 
coil  which  is  being  commutated  is  not  strong  enough  to  reverse 
the  current  in  the  time  of  commutation.  Sparking  will  therefore 
occur  in  generators  or  motors  at  heavy  load  when  the  armature 
m.m.f.  is  so  great  that  it  wipes  out  the  field  under  the  pole  tip  or 
weakens  it  to  such  an  extent  that  it  cannot  produce  the  required 
commutating  e.m.f.  Motors  will  also  spark  at  high  speed  since 
the  time  of  commutation  is  reduced,  especially  when  the  high 
speed  is  produced  by  field  weakening. 

Take  for  example  a  shunt  motor  rated  at  normal  speed  as 
10  hp.,  110  volts,  80  amp.,  400  r.p.m.  and  suppose  the  temperature 
rise  to  be  50°C.  If  the  motor  is  operated  at  half  speed  of  200 
r.p.m.  by  reducing  the  impressed  voltage  to  half,  the  rating  may 
be  taken  as  5  hp.,  55  volts,  80  amp.,  but  the  temperature  rise 
will  be  greater  than  before  because  the  armature  copper  loss  is  the 
same,  the  field  copper  loss  is  the  same  and  the  iron  and  friction 
losses  are  less  due  to  the  low  speed,  but  the  ventilation  is  only 
about  half  as  good  as  before. 

When  operated  at  twice  full  speed,  produced  by  field  weaken- 
ing, the  rating  may  be  taken  as  10  hp.,  110  volts,  80  amp., 
but  the  temperature  rise  will  be  less  than  before,  because  the 
armature  copper  loss  is  the  same,  the  field  copper  loss  is  reduced 
to  about  one-quarter  of  its  normal  value  and  the  iron  and  friction 
losses  are  increased,  but  the  ventilation  is  very  much  improved. 
The  rated  output  of  the  machine  for  normal  temperature  rise 
might  be  increased  but  due  to  the  higher  speed  and  consequent 
reduced  time  of  commutation  sparking  would  occur. 

211.  Temperature  Limits  (Standardization  Rules  A.  I.  E.  E. 
1914). — The  capacity  of  a  machine  is  limited  by  the  maximum 
temperature  at  which  the  materials  in  the  machine,  especially 
those  employed  for  insulation,  may  be  operated  for  long  periods 
without  deterioration.  When  the  safe  limits  are  exceeded, 
deterioration  is  rapid.  The  insulating  materials  become  per- 
manently damaged  by  excessive  temperature,  the  damage 
increasing  with  the  length  of  time  that  the  excessive  temperature 
is  maintained  and  with  the  amount  of  excess  temperature  until 
finally  the  insulation  breaks  down. 

The  actual  temperature  attained  in  the  different  parts  of  a 


254  ELECTRICAL  ENGINEERING 

machine  and  not  the  rises  in  temperature  affect  the  life  of  the 
insulation  of  the  machine.  The  safe  operating  temperatures  of 
the  various  parts  of  a  machine  are  often  expressed  in  terms  of  the 
allowable  temperature  rise  above  the  temperature  of  the  sur- 
rounding air.  The  temperature  of  this  air  is  called  the  ambient 
temperature  and  40°C.  is  taken  as  its  standard  value.  The 
allowable  temperature  rise  plus  40°C.  gives  the  maximum  allow- 
able temperature. 

A  machine  may  be  tested  at  any  convenient  ambient  tem- 
perature but  the  permissible  rises  of  temperature  must  not  exceed 
those  given  in  column  2  of  the  table  on  page  255. 

As  it  is  usually  impossible  to  determine  the  maximum  tem- 
perature attained  in  insulated  windings,  it  is  convenient  to  apply 
a  correction  to  the  measured  temperature  to  cover  the  errors  due 
to  the  fallibility  in  the  location  of  the  measuring  devices  as  well 
as  inherent  inaccuracies  in  measurement  and  method. 

The  two  most  usual  methods  of  measuring  temperature 
rises  are:  (1)  by  thermometer,  and  (2)  by  the  increase  of 
resistance. 

1.  Thermometers  are  applied  to  the  hottest  accessible  parts 
of  the  machine.     In  this  case  the  hottest  spot  temperature  for 
windings  should  be  estimated  by  adding  15°C.  to  the  highest 
temperature  observed. 

2.  The  measurement  of  the  increase  of  temperature  by  increase 
of  resistance  is  only  applicable  to  windings  of  comparatively 
high  resistance.     The  resistance  is  measured  before  operation 
at  the  ambient  temperature  and  again  after  operation.     The 
rise  of  temperature  can  then  be  calculated  as  shown  in  Art.  84. 
A  hot-spot  correction  of  10°C.  should  be  added. 

The  following  table  gives  the  limits  for  the  hottest  spot  tem- 
peratures of  insulations.  The  permissible  limits  are  indicated 
in  column  1  of  the  table.  The  limits  of  temperature  rise  per- 
mitted under  rated  load  conditions  are  given  in  column  2  and  are 
found  by  subtracting  40°C.  from  the  figures  in  column  1.  What- 
ever be  the  ambient  temperature  at  the  time  of  the  test,  the  rise 
of  temperature  observed  must  never  exceed  the  limits  in  column 
2  and  the  highest  temperature  must  never  exceed  the  limits  given 
in  column  1. 

Table  of  hottest  spot  temperatures  and  corresponding  per- 
missible temperature  rises. 


DIRECT-CURRENT  MACHINERY 


255 


Class 

Description  of  insulation 

Column  1, 
highest 
permissible 
temperature 
for  the  hot 
spot, 
degrees  C. 

Column  2, 
highest 
permissible 
temperature 
rise  for  the 
hottest  spot 
above  40°C. 

A, 
A, 
B 

C 

Cotton,  silk,  paper  and  other  fibrous  materials 
not  so  treated  as  to  increase  the  thermal 
limit.  

95 
105 

125 
No  limit  i 

55 
65 

85 
s  specified 

Similar  to  AI  but  treated  or  impregnated  or 
permanently  immersed  in  oil  and  including 
enameled  wire.  .  .  .  ... 

Mica,  asbestos  or  other  material  capable  of  re- 
sisting high  temperatures  in  which  any  class 
A  material  or  binder,  if  used,  is  for  structural 
purposes  only  and  may  be  destroyed  without 
impairing  the  insulating  or  mechanical 
qualities.  .  

Fireproof  and  refractory  materials  as  mica, 
porcelain  etc 

212.  Temperature  of  Commutators. — The  observable  tempera- 
ture shall  in  no  case  be  permitted  to  exceed  the  values  given  in 
the  table  above  for  the  insulation  employed,  either  in  the  com- 
mutator or  in  any  insulation  whose  temperature  would  be  affected 
by  the  heat  of  the  commutator.  For  commutators  so  constructed 
that  no  difficulties  from  expansion  can  occur,  the  following  tem- 
perature limits  have  been  suggested. 


Current  per  brush  arm 
200  amp.  or  less 
200  to  900  amp. 

900  amp.  and  over 


Maximum  permissible  temperature 

130°C. 

130°C.  less  5°  for  each  100  amp.  in- 
crease above  200 
95°C. 


213.  Temperature  of  Cores. — The  temperature  of  those  parts 
of  the  iron  core  in  contact  with  insulating  materials  must  not 
exceed  the  limits  of  temperature  and  temperature  rise  permitted 
for  those  materials. 

214.  Temperature   of   Other  Parts. — Other   parts    (such   as 
brush  holders,  brushes,  bearings,  pole  tips,  cores,  etc.),  whose 
temperature  does  not  affect  the  temperature  of  the  insulating 
material,  may  be  operated  at  such  temperatures  as  shall  not  be 
injurious  in  any  respect.     But  no  part  of  continuous  duty  machin- 


256 


ELECTRICAL  ENGINEERING 


ery  subject  to  handling  in  operation,  such  as  brush  rigging,  shall 
have  a  temperature  in  excess  of  100°C. 

215.  Ventilation. — The  increase  in  output  per  pound  of  active 
material  in  modern  machines  has  been  largely  due  to  improved 
methods  of  ventilation. 

Inlets  of  sufficient  size  and  proper  location  are  provided;  the 
cool  air  is  drawn  in  either  by  natural  suction  or  by  fans  placed  on 
the  rotating  member;  and  the  heated  air  is  expelled  through 
outlets  in  such  a  direction  that  it  will  be  thrown  completely 
away  from  the  machine. 

The  air  ducts  through  the  active  material  are  of  two  kinds, 
radial  ducts,  Fig.  242,  and  axial  ducts,  Fig.  243.  They  must  be 
of  sufficient  section  to  carry  the  required  amount  of  air  and  must 
have  sufficient  surface  to  allow  the  heat  to  pass  from  the  copper 
and  iron  to  the  air. 

J  _      I. 


FIG.  242  — Axial  ducts. 


FIG    243.— Radial  ducts. 


About  100  cu.  ft.  of  air  per  minute  per  kilowatt  lost  should  be 
provided. 

Care  must  be  taken  that  the  end  bells  or  overhanging  frame 
do  not  deflect  the  hot  air  back  into  the  machine. 

A  generator  when  driven  by  a  belt  may  run  cool  due  to  windage 
from  the  belt,  whereas,  if  it  is  direct-connected  to  a  motor  or 
steam  engine  it  may  run  hot  due  to  poor  ventilation  or  due  to 
the  transfer  of  hot  air  from  the  driving  machine. 

Machines  of  short  length  and  large  diameter  are  easy  to  cool. 

216.  Semi-enclosed  and  Totally  Enclosed  Machines. — When 
a  machine  is  partially  enclosed  it  is  more  difficult  to  get  rid  of  the 
heat  due  to  the  losses  and  unless  a  more  efficient  system  of  ven- 
tilation is  provided  the  output  must  be  decreased  15  or  20  per 
cent,  to  keep  within  the  allowable  limits  of  temperature  rise. 


DIRECT-CURRENT  MACHINERY  257 

When  totally  enclosed  the  rating  must  be  still  further  reduced. 
For  the  same  temperature  rise  a  totally  enclosed  machine  can  give 
about  70  per  cent,  of  the  output  of  an  open  machine  if  the  speed 
is  increased  about  20  per  cent.  The  reduction  of  output  reduces 
the  copper  losses  and  the  increase  of  speed  decreases  the  flux  per 
pole  and  so  decreases  the  iron  losses  but  not  in  direct  proportion. 


17 


CHAPTER  VIII 
DESIGN  OF  A  DIRECT-CURRENT  GENERATOR 

217.  Symbols. — The  subscripts  used  are  c  for  core,  g  for  air  gap,  t  for 
teeth,  p  for  pole  and  y  for  yoke. 

A  =  sectional  area  in  square  inches. 
AT  =  ampere-turns. 
AT  g  =  ampere-turns  per  pole  for  the  gap. 

B  =  flux  density  in  lines  per  square  inch. 
Bag  =  actual  gap  density. 
Bat  =  actual  tooth  density. 
Bg  =  apparent  gap  density. 
Bt  =  apparent  tooth  density. 
Da  =  external  armature  diameter. 
De  =  commutator  diameter. 

I  =  load  current. 
Ic  =  current  per  conductor. 
Lc  =  axial  length  of  core. 
Lg  =  gross  iron  in  the  frame  length. 
Ln  =  net  iron  in  the  frame  length. 

S  =  number  of  commutator  segments. 
Tf  =  field  turns  per  pole. 

Z  =  total  conductors  on  the  armature. 

d  =  slot  depth. 

p  =  number  of  poles. 
pi  =  number  of  paths  through  armature. 

q  =  ampere  conductors  per  inch. 

s  =  slot  width. 

t  =  tooth  width  at  top. 

5  =  air-gap  clearance. 

17  =  efficiency. 

X  =  slot  pitch. 

v  =  leakage  factor. 

T  =  pole  pitch. 

0  =  magnetic  flux. 
<t>a  =  useful  flux  per  pole. 

^  =  per  cent,  pole  enclosure. 

The  meaning  of  other  symbols  used  will  be  explained  in  the  context. 
Figs.  244,  245,  246,  247,  252,  257  are  reproduced  from  the  "Standard 
Handbook,"  Section  8,  by  A.  M.  GRAY. 

258 


DESIGN  OF  A  DIRECT-CURRENT  GENERATOR    259 

218.  Design  of  Direct-current  Machinery. — The  design  of 
electrical  machinery  is  based  on  a  number  of  formulae  or  equa- 
tions which  can  be  very  easily  derived  and  on  a  number  of  em- 
pirical relations  obtained  from  experience,  which  are  usually 
expressed  in  the  form  of  curves  or  limiting  values. 

Following  are  the  most  important  of  these  fumiamental  rela- 
tionships and  limits.  They  are  indicated  by  Roman  numerals 
for  purposes  of  reference. 

/.  The  e.m.f.  equation: 

generated  e.m.f.  =   8  =  Zn^g  £-  10~8  volts  (Art.  142), 
or 

c 

useful  flux  =  0ff  =  -  lines<  (270) 

Zn  "  10~8 

II.  The  output  equation: 

output  in  watts  =   8/  =  ( —  )(<t>op)n  X  10~8 

10~8 


60 


r.p.m.)         (271) 


=  a  constant  X  electric  loading  X  magnetic  loading  X  r.p.m. 
ZIC  =  total  ampere  conductors  on  the  armature  is  called  the 

electric  loading. 
4>gp  =  total  flux  crossing  the  gaps  under  the  poles  is  called  the 

magnetic  loading. 

A  large  electric  loading  requires  a  large  amount  of  copper  in 
the  machine,  while  a  large  magnetic  loading  requires  a  large 
amount  of  iron. 

The  output  equation  can  be  written  in  a  more  easily  applicable 
form: 

10~8 
watts  =  --(TrDaq)(Bg^TLcp)  (r.p.m.) 


'QQ-vDoQ  Bat~Lep  X  r.p.m. 


60X108 
or 


r.p.m. 


Da2Lc  qBg\l/  X  r.p.m. 

»   V    1H7 

(272) 


260 


ELECTRICAL  ENGINEERING 


where, 

Da2Lc  =  volume  of  the  active  material  in  the  armature. 
Bg  =  apparent  density  in  the  air  gap. 
\f/  =  per  cent,  pole  enclosure. 

ZI 

q  =  —  fp=  ampere   conductors   per  inch   of  the    armature 
irJJa 

periphery. 

///.  Flux  Densities.  —  The  Qux  density  in  the  air  gap  is  limited 
by  the  maximum  allowable  flux  density  in  the  roots  of  the  teeth, 
which  is  dependent  on  the  frequency  of  the  reversals  of  the 
magnetism. 

The  approximate  limits  of  the  maximum  tooth  density  are 
150,000  lines  per  square  inch  for  30  cycles  per  second  and  125,000 
lines  for  60  cycles.  The  frequency  is  given  as 

/  =  ~  cycles  per  second. 

Zi 

Having  fixed  the  density  at  the  roots  of  the  teeth  the  air-gap  den- 
sity will  depend  on  the  diameter  of  the  armature. 


Gap  Density  In 

§ 


&OX10* 


0         20         40         60        80        100 
Armature  Diameter  In  Inches  =  D 


Ampere  Conductors  per  Inch  =  Q 

o  §  §  §  §  i  §  i 

N 

1'  10 

s 

^~~ 

^=± 

,---' 

-  —  ' 

,  • 

.  — 

.>- 

-H 

^f 

X* 

-Scale 

B 

s 

j 

/ 

8 

calJ  B 

2000 

4000 

»; 

UO 

JO          1 

i 

ot 

(1 

4C 

1 

0 

) 

HOj 

-Ttn 

I 

H. 

•M 

^ 

s^ 

.  — 

i  u 

^ 

X 

/* 

'Be 

ale 

A 

1 

"/ 

/ 

s 

/ 

/ 

0 

j 

/ 

•i 

a 

1 

tt_ 

J 

)               L- 

i 

/ 

a.i 

jf 

__ 

1 

^  — 

^—  • 

' 

ti 

0           ( 

Scale 

A 

>          20        40         60         80       10 
Kilowatts 

FIG.  245. 

>         20         40          CO          30        10 
Watts 
R.P.M. 

FIG.  246 

FIG.  244. 

Fig.  244  shows  the  relation  between  Bg  and  Da  for  a  frequency 
of  30  cycles  per  second.  Lower  values  of  Bg  should  be  used  for 
higher  frequencies. 

The  ordinary  flux  densities  in  other  parts  of  the  magnetic 
circuit  are: 

Flux  density,  lines  per  square 
inch,  depending  on  the  frequency 

70,000-100,000 


Armature  core 
Pole  (cast  steel) 
Yoke  (cast  iron) 
Yoke  (cast  steel) 


90,000-  95,000 
35,000-  40,000 
70,000-  80,000 


DESIGN  OF  A  DIRECT-CURRENT  GENERATOR    261 


IV.  Ampere    Conductors    per    Inch.  —  The    value    of    q    the 
ampere  conductors  per  inch  of  armature  periphery  is  limited 
partly  by  heating  and  partly  by  commutation  and  it  depends 
on   the  output   of  the  machine.     Fig.  245  shows  the  relation 
between  q  and  kilowatt   output  for  machines  in   satisfactory 
operation. 

V.  Electric  Loading.  —  For  the  most  economical  construction 
there  is  a  more  or  less  fixed  relation  between  the  electric  and  the 
magnetic  loading. 

Therefore,  referring  to  equation   (271),  it  is  seen  that  the 
electric  loading,  or  the  total  ampere  conductors  on  the  armature, 

.     watts 

depends  on  the  ratio  -      — 
r.p.m. 

The  relation  between  ampere  conductors,  ZIC  and  watts  per 
revolution  is  shown  in  Fig.  246. 

VI.  Current  Density.  —  The  current  density  used  in  the  arma- 
tures of  direct-current  generators  varies  from  2,000  to  3,000  amp. 
per  square  inch,  that  is,  600  to  400  circ.  mils  per  ampere.     It  is 
limited  by  the  allowable  temperature  rise.     The  value  to  be  used 
on  a  given  machine  depends  on  the  specific  electric  loading  q 
and   on   the   peripheral   speed.     Fig.    247   shows   the   relation 

,  ,  .    ampere  conductors  per  inch       . 

between  the  ratio  -  —  £  —  -,  -  ^—  -  and  the  peripheral 

circular  mils  per  ampere 

speed  in  feet  per  minute  for 
a  temperature  rise  of  40°C. 

VII.  Choice  of  Number  of 
Poles.—  The  number  of  field 
ampere-turns  per  pole  is  usu- 
ally   about     50     per    cent. 
greater   than   the  armature 
ampere-turns    per    pole.     If 
the  number  of  poles  is  too 
small  the  required   ampere- 
turns  per  pole  will  be  large 
and  the  poles  must  be  made 
long  to  carry  the  windings. 

The  number  of  poles  may  be  fixed  approximately  by  the  two 
relations  : 


Z*4 

/ 

§9  n 

& 

Y 

ko 

I 

,<?> 

7 

5 

B  1  JJ 

(V  < 

/ 

^ 

° 

/ 

o  10° 
O 

\ 

/ 

\> 

/^ 

PV 

. 

/ 

/ 

V 

Sv 

/ 

/ 

/ 

x 

^ 

/ 

f 

/ 

a 

/ 

/ 

•^ 

fS 

ftt° 

•$ 

^ 

'/ 

c 

,-- 

2» 

I™ 

^~~- 

7{     o 

0        ( 

1 

\ 

' 

4 

1 

l 

a  8° 


Peripheral  Velocity  in  Ft.  per  Mio. 

FIG.  247. 


(a)  Ratio 


Ues  between  L1  and  L7  usually- 


ength 
(6)  Armature  ampere-turns  per  pole  should  not  exceed  7,500. 


262  ELECTRICAL  ENGINEERING 

VIII.  Commutation. — To  provide  a  commutating  field,  the 
.    field  ampere-turns  per  pole  for  the  gap  and  teeth 

armature  ampere-turns  per  pole  ~  should  not 

be  less  than  1.2.  If  the  brushes  are  advanced  so  that  part  of  the 
armature  m.m.f.  is  demagnetizing  the  field  ampere-turns  must 
be  increased  by  the  amount  of  the  armature  demagnetizing  am- 
pere-turns. 

IX.  Reactance    Voltage. — The    formulae    for     the     reactance 
voltage  were  derived  in  Art.  195. 

ER  =  K  X  Lc  X  ZIC  X  r.p.m.  X  n  X  10~8  volts 

where  K  is  a  constant. 

For  a  full-pitch  multiple  winding  K  =  0.8,  and  for  a  short- 
pitch  multiple  winding  K  =  0.46. 

With  the  armature  strength  limited  as  indicated  in  VIII  above 
and  the  brushes  advanced  into  a  suitable  commutating  field, 
the  reactance  voltage  with  a  full-pitch  winding  should  not  ex- 
ceed 1.5  to  2  vol^s  for  shunt  machines.  In  compound-wound 
machines  values  up  to  3.5  volts  may  be  taken  care  of  in  certain 
cases.  However,  when  the  reactance  voltage  is  much  above  2.5 
volts  it  may  be  advisable  to  use  interpoles. 

With  short-pitch  windings  the  two  sides  of  the  short-circuited 
coil  are  not  in  commutating  fields  of  equal  strength  and  slightly 
lower  limits  should  be  used. 

With  series  windings  since  a  number  of  coils  are  short-circuited 
in  series  the  reactance  voltage  calculated  by  the  formula  for  the 
full-pitch  winding  should  be  kept  about  50  per  cent,  lower  than 
for  the  multiple  winding. 

X.  Slots. — The  ratio  -=— - — rVrr  varies  from  2.5  to  3.5. 

slot  width 

slot  width 
rheratl°  maximum  tooth  width  ==  l  Approximately. 

Slots  per  pole  should  be  greater  than  12  for  small  machines 
and  14  for  large  machines. 

XI.  Commutator. — The  commutator  diameter  is  from  60  to 
75  per  cent,  of  the  armature  diameter.     If  possible  the  peripheral 
speed  of  the  commutator  should  not  exceed  3,500  ft.  per  minute 
but  values  up  to  5,000  may  be  used  in  special  cases. 

XII.  Brushes. — The  brush  arc  should  not  cover  more  than 
three  segments  and  should  not  subtend  more  than  one-twelfth 
of  the  pole  pitch. 


DESIGN  OF  A  DIRECT-CURRENT  GENERATOR    263 


The  current  density  in  the  brushes  depends  on  the  kind  of 
brush  used.  For  ordinary  brushes  from  35  to  40  amp.  per  square 
inch  may  be  used  and  the  contact  resistance  drop  may  be  taken 
as  1  volt  per  brush.  This  gives  a  loss  of  35  to  40  watts  per  square 
inch  of  brush  contact,  which  is  satisfactory.  Where  brushes  with 
lower  contact  resistances  are  used  the  current  density  may  be 
higher. 

The  brush  pressure  is  from  IJ^j  to  2J^  Ib.  per  square  inch  de- 
pending on  the  service. 

The  coefficient  of  friction  between  the  brush  and  the  commuta- 
tor may  be  taken  as  0.3  for  ordinary  hard  carbon  brushes. 

219.  Magnetic  Leakage. — Since  there  is  no  material  through 
which  magnetic  flux  cannot  pass,  it  is  not  possible  to  confine  all 
the  flux  produced  in  a  generator  to  the  magnetic  circuit.  In 
practice  the  main  circuit  is  made  of  such  low  reluctance  that  only 
a  small  portion  of  the  flux  leaves  it. 


FIG.  248. — Leakage  flux  about  a 
bipolar  dynamo. 


FIG.  249. — Leakage  flux  about  a 
multipolar  dynamo. 


Figs.  248  and  249  show  the  leakage  fluxes  about  the  magnetic 
circuits  of  a  bipolar  and  multipolar  generator. 

The  principal  part  of  the  leakage  occurs  between  the  pole  tips 
because  the  m.m.f.  consumed  between  these  points  is  from  60  to 
80  per  cent,  of  the  total  m.m.f. ;  it  includes  the  m.m.f.  required  to 
drive  the  flux  across  the  two  gaps  and  through  the  teeth  and 
armature  core.  As  a  result  the  flux  passing  through  the  field 
poles  and  yokes  is  greater  than  the  flux  crossing  the  gap  into  the 
armature  by  an  amount  depending  both  on  the  mechanical  con- 
struction of  the  machine  and  on  the  load. 

The  dispersion  coefficient  or  leakage  factor  is  the  ratio  of  the 
flux  through  the  field  poles  to  the  flux  crossing  the  gap  into  the 
armature,  that  is,  the  ratio  of  the  total  flux  to  the  useful  flux; 
thus  the  leakage  factor  is 

<£  P°le 


264     '•;;.;  ELECTRICAL  ENGINEERING 

In  preliminary  calculations  the  following  values  of  leakage 
factor  may  be  used  : 

Four-pole  machines  up  to  10  in.  armature  diameter  ........  1  .25 

Multipolar  machines  from  10  to  30  in.  armature  diameter.  .  .  1.2 

Multipolar  machines  from  30  to  60  in.  armature  diameter.  .  .  1  .  18 

Multipolar  machines  over  60  in.  armature  diameter  ........  1  .  15 

Under  load  the  armature  exerts  a  m.m.f.  which  in  part  is 
demagnetizing  and  opposes  the  field  m.m.f.  and  in  part  is  cross- 
magnetizing  and  increases  the  reluctance  of  the  magnetic-  circuit 
(Art.  145).  Thus  under  load  a  greater  proportion  of  the  field 
m.m.f.  is  required  for  the  air  gaps,  teeth  and  armature  than  at  no 
load  and  the  leakage  flux  is  increased  in  proportion.  The  leakage 
factor  is  therefore  greater  under  load  than  at  no  load. 

In  the  case  of  the  flat-compound  generator,  designed  in  this 
chapter  armature  reaction  consumes  a  component  of  the  full- 
load  field  m.m.f.  =  2,117  ampere-turns  per  pole.  At  no  load 
the  m.m.f.  consumed  in  the  gap,  the  teeth  and  the  armature  core 
is  5,928  ampere-turns  per  pole  and  the  leakage  factor  is  taken  as 
1.18.  At  full  load  the  leakage  factor  is  increased  to 


220.  Design  of  a  Direct-current  Generator.  —  Determine  the 
dimensions,  and  characteristics  of  a  generator  of  the  following 
rating:  250  kw.,  250  volts,  400  r.p.m.  (Fig.  250). 


Armature  design. 
Watts       250,000 


_ 


Rrn    ~      400 

Ampere  conductors  =  ZIC  =  78,000,  Fig.  246. 

Ampere  conductors  per  inch  =  q  =  690,  Fig.  245. 

ZIC       78,000 

Armature  circumference  =  irDa  =  —  =     Ann     =  llo  m. 

5 

•I  -I  O 

Armature  diameter  =  Da  =  —  -  =  36  in. 

Apparent  gap  density  =  Bg  =  55,000,  Fig.  244. 
Pole  enclosure  =  \f/  =  0.7,  assumed. 
watts       60.8  X  107 

D»*L*  =      x  —      equation  (272)- 


250,000  60.8  X  107      _  v  _1  __  1t  . 

Frame  length  =  Lc  =       400      X  55j000  X  0.7  X  690  X  (36)  2  ~ 

Number  of  poles  =  p  =  8,  assumed. 

7  T         78  000 
Armature  ampere-turns  per  pole  =  -g-5  =  2x8   =  4,860. 


DESIGN  OF  A  DIRECT-CURRENT  GENERATOR    265 


TrDa          IT    X36 

Pole   pitch  =  r  = =  — g — 


14.13  in. 


Pole  pitch 


14.13 
11 


1.3. 


Frame  length       Lc 

Area  of  the  air  gaps  =  Ag  =  ^rLc  =  0.7  X  14.13  X  11  =  109  sq.  in. 

Useful  flux  per  pole  =  <£„  =  BgAg  =  55,000  X  109  =  6  X  106  lines. 

Winding  assumed  to  be  multiple,  short  pitch,  with  one  turn  per  coil. 

Reactance  voltage  =  ER  =  0.46  X  10"8  X  Lc  X  ZIe  X  r.p.m.  X  n, 
equation  (268)  =  0.46  X  10~8  X  11  X  78,000  X  400  X  1  =  1.58.     With  a 
full-pitch  winding  it  would  have  been  2.7  volts. 


FIG.  250. — Magnetic  circuit. 

Number  of  path  in  multiple  between  terminals  =  pi  =8. 
Total  conductors  on  the  armature  = 

7       S_  250  X  108 

Z  =  =  77^ o  =  624. 


Conductors  per  slot  =  4,  assumed. 
Number  of  slots  =  -7-  =  156. 

156 
Slots  per  pole  =  —  -  =  19.5. 

o 

Peripheral  velocity  of  the  armature  in  feet  per  minute  = 

36 


TT  X        X  400  =  3,800. 


Ampere  conductors  per  inch 
Circular  mils  per  ampere     for  4°  C'  rlse  =  L4'  Fl«'  247' 

Circular  mils  per  ampere  =  ^-7  =  y-j  =  494. 


266 


ELECTRICAL  ENGINEERING 


Total  load  current 


watts 
volts 


61,750  circ.  mils  =  0.0485  sq.  in. 


~  r         /       1,000 

Current  per  conductor  =  lc  =  —  =  — ^—  =  125  amp. 

Section  of  conductor  =  125  X  494 

Slot  Pitch  -  *  -  numbefo'f  slots  '  ^  "  °'725  in' 
Slot  width  =  0.5  X  slot  pitch,  assumed  =  0.5  X  0.725  =  0.3625. 
Conductor  dimensions  =  width  X  depth  =  x  X  y  =  0.0485  sq.  in. 
Thickness  of  insulation  in  width  =  0.112  in.   (Fig.  251.) 
Allowance  for  clearance  =  0.040  in. 


}.375->| 


nductor  O.l  *  0.5 
Half  Lapped  Tape  0.012 
Half  Lapped  Tape  0.012 
Empire  Cloth  0.010 
Papw  0.010 


FIG.  251. — Details  of  slot  and  insulation. 


Width  of  two  conductors  =  2z  =  0.3625  -  0.152  =  0.2105. 

Width  of  conductor  =  x  =  0.105  =0.1  in. 

Slot  width  =s  =  2X0. 1+0. 112+  0.040  =  0.352  =  0.35  in. 

Depth  of  conductor  =  y  =    '    ^      =  0.485  =  0.5  in. 

Circular  mils  per  ampere  =  494  X  —  -  =  510. 

0.0485 
4      106 
Amperes  per  sq.  inch  =  -  X^rx  =  2,500. 

TT          OlU 

Thickness  of  insulation  in  slot  depth  =  0.216  in. 

Thickness  of  wedge  =  0.2. 

Depth  of  two  conductors  =  1.0. 

Allowance  for  clearance  =  0.040. 

Depth  of  slot  =  1.456  =  1.45  in. 

Maximum  tooth  width  =  t  =  \  —  s  =  0.725  —  0.35  =  0.375  in. 

Diameter  at  bottom  of  slots  =  (36  -  2  X  1.45)  =  33.10  in. 

vx    00    -I  Q 

Minimum  tooth  width  =        „*' 0.35  =  0.67  -  0.35  =  0.32  in. 


Tooth  taper  =  *  =  ^ 


x. 
faun. 


156 
0.375 
0.32 


1.17. 


Number  of  vent  ducts  in  the  core  =  3  -  <Ke  in.  wide. 
Gross  iron  in  the  frame  length  =  La  =  Lc  —  vent  ducts  = 

(11  -  3  X  0.3125)  =  10.06  in. 
Stacking  factor  for  armature  punchings  =  0.9. 


DESIGN  OF  A  DIRECT-CURRENT  GENERATOR    267 

Net  iron  in  the  frame  length  =  Ln  =  0.9Lff  =  0.9  X  10.06  =  9.05  in. 

number  of  teeth 
Number  of  teeth  under  one  pole  =  ^  X  -         ~^~       -  =  0.7  X 

156        i*fi 
-g-     =  13.6. 

Maximum  tooth  area  per  pole  =  13.6  X  0.375  X  9.05  =  46  sq.  in. 
Minimum  tooth  area  per  pole  =  13.6  X  0.32  X  9.05  =  39.5  sq.  in. 
Flux  in  the  teeth  =  <fr  =  (f>0  =  6  X  106  lines. 

6  X  106 
Apparent  minimum  flux  density  in  the  teeth  =  —  ^  —  =  131,000  lines 

per  square  inch,  Art.  221. 

6  X  106 
Apparent  maximum  flux  density  in  the  teeth  =  =  152,000  lines 


per  square  inch. 
,       p  X  n       8  X  400 
Frequency  of  the  reversals  of  magnetism  =  /  =  —  ^  —  "  =  2  X  60    = 

27  cycles  per  second. 
Flux  density  in  the  core  =  85,000  lines  per  square  inch,  assumed. 

Flux  in  the  core  =  <t>c  =  |ff  =  6  \  ^  =  3  X  106. 

3  X  106 
Core  area  =  -oe  Ann    =  35.4  sq.  in. 

OO,UUU 

core  area       35.4 
Depth  of  iron  below  slots  =  —  y  --  =  Q-Q^  =  3.92  in. 

Internal  diameter  of  armature  =  J36  -  2(1.45  +  3.92)}  =  25.16  =  25  in. 

QC  _  25 
Depth  of  iron  below  slots  =  -  ^  ---  1.45  =  4.05  in. 

Core  area  =  Ac  =  4.05  X  9.05  =  36.7  sq.  in. 

6  X  106 
Flux  density  in  the  core  =  Bc  =  0  ^  QA  n  =  82>000  lines  per  square  inch. 

L    /\  oO.  I 

221.  Flux  in  the  Air  Spaces  between  the  Teeth.  —  If  the  flux 
density  in  the  teeth  is  above  100,000  lines  per  square  inch  it  is 
necessary  to  take  account  of  the  fact  that  the  path  through  the 
teeth  is  paralleled  by  an  air  path  consisting  of  the  slots,  the  vent 
ducts  and  the  insulation  between  punehings.  This  path  has 
usually  a  much  larger  section  than  the  path  through  the  teeth 
and  consequently  at  high  densities  where  the  permeability  of  the 
iron  is  low  it  will  carry  a  considerable  part  of  the  flux. 

In  the  case  of  the  generator  designed: 

Maximum  tooth  area  per  pole  =  46  sq.  in. 
Gap  area  =  109  sq.  in. 

Area  of  path  through  slots  and  vent  ducts  =  63  sq.  in. 
Minimum  tooth  density  =  131,000  lines  per  square  inch. 
Ampere-turns  per  inch  for  this  density  =  800. 
Flux  density  in  the  parallel  air  path  =  800  X  3.2. 
Flux  in  the  parallel  path  =  800  X  3.2  X  63  =  160,000. 
The  average  ampere-turns  per  inch  for  the  teeth  will  be  greater  than  800 
due  to  the  tooth  taper  (Art.  222)  and  it  will  be  better  to  assume  that  the 


268 


ELECTRICAL  ENGINEERING 


flux  carried  by  the  air  path  is  somewhat  greater  than  160,000,  assume 
200,000,  then  flux  in  the  teeth  =  6  X  106  -  200,000  =  5.8  X  106.. 
Maximum  flux  density  in  the  teeth  =  147,000  lines  per  square  inch. 

222.  Effect  of  Tooth  Taper.— When  the  diameter  of  the  arma- 
ture is  small  or  the  slots  are  deep,  the  flux  density  at  the  roots 
of  the  teeth  is  much  greater  than  that  at  the  tops.  If  the  ampere- 
turns  for  the  teeth  are  calculated  by  using  the  maximum  density, 
the  result  is  too  large,  while  if  the  minimum  density  is  used  the 
result  is  much  too  small.  The  curves  in  Fig.  252  have  been 

calculated  for  various  tooth  tapers,  k  =          ',  and  give  the  average 

ampere-turns  per  inch.     Corresponding  to  the  actual  tooth  den- 
sities at  the  root. 


icono* 


100      120      140      160      180     200 
Turns  per  In.  •* 

FIG.  252. — Magnetization  curves  for  tapered  teeth. 


With  a  flux  density  of  147,000  lines  per  square  inch  at  the  root 
and  a  tooth  taper  k  =  1.17  the  ampere-turns  per  inch  required  = 
1,065. 

223.  Ampere -turns  per  Inch  for  an  Air  Path. — When  a  flux 
density  B  lines  per  square  inch  is  to  be  produced  in  an  air  path, 
the  ampere-turns  per  inch  required  may  be  found  as  follows : 


Taking  a  path  1  in.  long  and  1  sq.  in.  in  section  <£  =  J5;  I  =  1 


DESIGN  OF  A  DIRECT-CURRENT  GENERATOR    269 


in.  =  2.54  cm.,   A  =  1   sq.   in.  =  (2.54)2  sq.   cm.,  ju  =  1;  and 
substituting 

0.4  X  3.14  (nl) 


B  = 


2.54 


(2.54) 2  X  1 


and  the  ampere-turns  per  inch. 

2  54 

(nl)  per  inch  =  B  X 


0.4  X  3.14  X  (2.54) 2      ™™~      3.2*    (273) 
224.  Air  Gap. — The  air  gap  is  the  most  important  part  of  the 

magnetic  circuit  as  it  requires  the  largest  proportion  of  the  field 

m.m.f. 

The  section  of  the  air  gap  is  Ag  =  \f/rLc  and  the  apparent  flux 

density  is  Bg  =  -/-  =  ~rj~'     This  would  be  the  correct  density 

if  the  flux  were  uniformly  distributed,  but  due  to  the  presence 
of  the  slots  and  vent  ducts  the  whole  space  is  not  utilized  and  the 
actual  flux  density  is  Bag  =  CBg,  where  C  is  a  constant  called 
the  Carter  coefficient  and  depends  on  the  tooth  width,  slot  width 
and  gap  length.  Referring  to  Fig.  253, 


C  = 


t+fs 


(274) 


and  the  values  of  /  which   depend   principally  on   the  ratio 

slot  width       s  ,  x  .      .  , 

gap  len~th  =  5  mav        obtained  from  the  curve. 


FIG.  253. — Carter's  coefficient. 

The  ampere-turns  per  inch  required  for  the  air  gap  =  0.3132  X 

Z?  /nr  o 

T"»  Jj  Q,Q  vyX)/7 

B  ag  —  -zr^  —  -5-77  (Art.  223) ;  and  the  ampere-turns  per  pole  for 

O«*j  O.w 

the  gap 

AT°  =  ~^6'  (275) 

226.  Length  of  the  Air  Gap. — 

Armature  ampere-turns  per  pole  =  4,860. 

Ampere-turns  per  pole  (gap  +  teeth)  =  AT0+t  =  1.2  X  4,860  =  5,832. 


270  ELECTRICAL  ENGINEERING 

Ampere-turns  per  pole  for  the  teeth  =  ATt  =  1,550. 

Ampere-turns  per  pole  for  the  gap  =  ATg  —  4,282. 

gX  5000  X» 


Gap  length  =  5  =  0.22. 

s   _  0.35  _ 

I  ~  022  ~  L59' 

/  =  0.76,  Fig.  253. 

t  +   s  0.375  +  0.35 


*  +/s       0.375  +0.76  X0.35 
'Ampere-turns  for  the  gap 

_  CB0d  _  1.13  X  55,000  X  0.22 
~"A7(r~~!£2~  3.2 

226.  Design  of  Poles  and  Yoke.  — 

Density  in  the  pole  core  =  95,000  lines  per  square  inch  (assumed). 

Leakage  coefficient  =  v  =  1.18,  Art.  219. 

Flux  in  the  pole  =  <f>p  =  6  X  106  X  1.18  =  7.08  X  106  lines. 

Section  of  the  pole  core  =      Q  5  QQQ  —  =  74.6  sq.  in. 

Assume  a  round  pole  10  in.  in  diameter. 

Section  of  pole  =  Ap  =  -^-  X  102  =  78.5  sq.  in. 

Flux  density  in  the  pole  core  =  Bp  =  ^^  —  =  90,000  lines  per 

square  inch. 

The  field  ampere-turns  per  pole  are  approximately  50  per  cent,  greater 
than  the  armature  ampere-turns  per  pole  =  1.5  X  4,860  =  7,290;  about 
1,000  ampere-turns  can  be  placed  on  1  in.  of  winding  space  and,  there- 

7  290 
fore,  the  space  for  the  shunt  coil  is  -TQQQ  =7.3  in.  ;  adding  30  per  cent. 

to  this  to  take  care  of  the  series  field  the  total  winding  space  is  10  in.  and 
allowing  1  in.  for  the  pole  face,  the  length  of  the  pole  is  lp  =  11  in. 

Yoke  material  =  cast  iron. 

Yoke  density  =  40,000  lines  per  square  inch  (assumed). 

Flux  in  yoke  =  7'082X  10*  =  3.54  X  106. 

3  54  X  106 
Section  of  the  yoke  is      4Q  QQQ      =  88-5  S<1-  in->  take  Av  =  95  scl-  in- 

S  54  X  106 
Density  in  the  yoke  =  Bv  =  -     g~     -  =  37,300. 

227.  Determination  of  the  No-load  Saturation  Curve.  —  Fig. 
250  shows  the  dimensions  of  the  magnetic  circuit  of  the  eight- 
pole,  250-kw.,  250-volt,  400-r.p.m.  generator.     It  is  required  to 
obtain  the  no-load   saturation,  curve  for  this  machine.     The 
armature  has  a  multiple  winding  with  624  conductors. 


DESIGN  OF  A  DIRECT-CURRENT  GENERATOR    271 


The  voltage  generated  in  the  armature  is 


8  =  Zn<l>g    -  10~8  =  624  X          X  </>,  X  f  10~8  volts, 
PI  uU  o 

and  therefore 

4>g  =  24,0008. 

To  produce  the  rated  voltage,  250  volts,  a  flux  is  required  in 
the  gap,  <f>g  =  24,000  X  250  =  6  X  106  lines. 

The  leakage  factor  may  be  assumed  to  be  1.18,  Art.  219,  so 
that  the  flux  per  pole  is 

<j>p  =  1.1800  =  1.18  X  6  X  106  =  7.08  X  106. 
The  sections  and  lengths  of  the  various  parts  of  the  magnetic 
circuit  may  be  obtained  from  the  design  sheet  or  from  the  sketch. 


0       10    20    30     40    50     60    70     80    90    100 
Ampere  Turns  per  Inch 


25      50     75    100   125  150  175  300  225  250 
Ampere  Turus  per  lucb 


FIG.  254. — Magnetization  curves. 

Yoke. 

Material  =  cast  iron. 

Section  =  Av  =  95  sq.  in. 

Length  of  magnetic  path  =  ly  =  12  in.,  Fig.  250. 

Flux  in  yoke  =  j>y  =  -£  =  -  — ^ =  3.54  X  106  lines. 

3  54  X  106 
Flux  density  =  By  =— — ^ =  37,300  lines  per  square  inch. 

Ampere-turns  per  inch  =  65,  Fig.  254. 

Ampere-turns  per  pole  for  the  yoke  =  ATy  =  65  X  12  =  780. 
Pole. 

Material  =  cast  steel. 
Section  =  ,AP  =  78.5  sq.  in. 
Length  =  11  in. 
Flux  =  <t>P  =  7.08  X  106  lines. 

Flux  density  =  Bp  =  -^—^TK —  =  90,000  lines  per  square  inch. 


272  ELECTRICAL  ENGINEERING 

Ampere-turns  per  inch  =  48,  Fig.  254. 
Ampere-turns  per  pole  for  the  pole  core  =  48  X  11  =  528. 
Air  gap. 

Section  =  Ag  =  109  sq.  in. 
Length  =  6  =  0.22  in. 
Flux  =  fa  =  6  X  106  lines. 

i  A   \S   1  A6 

Apparent  flux  density  =  Bg  =  -~  =  — ^ —  =  55,000  lines  per  sq.  inch. 

Carter  coefficient  C  =  1.13. 

Actual   flux  density  =  Bag  =  CBa  =  1.13  X  55,000  =  62,150   lines   per 
square  inch. 

Ampere-turns  per  inch  =  -^  =     0'  0      =  19,400. 

O.6  O.« 

Ampere-turns  per  pole  for  the  gap  =  19,400  X  0.22  =  4,270. 

Teeth. 

Material  =  sheet  steel. 
Maximum  section  =  46  sq.  in. 
Minimum  section  =  At  =  39.5  sq.  in. 
Length  of  magnetic  path  =  It  —  1.45  in. 
Flux  =  fa  =  6  X  106  lines. 

6  X  106 
Apparent  maximum  flux  density  =     on  g      =  152,000  lines  per  square 

inch. 

Apparent  minimum  flux  density  =  — -r^ —  =  131,000    lines  per  square 

inch. 

Actual  maximum  flux  density  =  147,000  lines  per  square  inch,  Art.  221. 

Tooth  taper  =  k  =  1.17. 

Ampere-turns  per  inch  =  1,065,  Fig.  252. 

Ampere-turns  per  pole  for  the  teeth  =  ATt  =  1,065  X  1.45  =  1,550. 

Armature  core. 

Material  =  sheet  steel. 
Section  =  Ac  =  36.7  sq.  in. 
Length  =  lc  =  6  in.,  Fig.  250 

Flux  =  <t>c  =  f  =  ^i?8  =  3  X  106  lines. 

3  X  106 

Flux  density  =  Bc  =     oa  -      =  82,000  lines  per  square  inch, 
oo./ 

Ampere-turns  per  inch  =  18. 

Ampere-turns  per  pole  for  the  core  =  ATe  =  18  X  6  =  108. 

The  ampere-turns  per  pole  required  for  a  voltage  of  250  volts  at  no  load  is 
ATV  +  ATP  +  AT0  X  ATt  +  ATC  =  780  +  528  +  4,270  +  1,550  +  108 
=  7,236. 

These  results  are  tabulated  below  and  also  the  results  of  similar 
calculations  made  for  225  volts  and  275  volts. 

From  these  results  the  saturation  curve  for  the  machine  is 
plotted  in  Fig.  256.  Volts  vs.  field  ampere-turns  per  pole. 


DESIGN  OF  A  DIRECT-CURRENT  GENERATOR    273 


No-load  voltage  =  8 

225 

250 

275 

Useful  flux  per  pole  =  4>o 

5.4  X  10« 

6  X  10« 

6.6  X  106 

Leakage  factor 

1.18 

1.18 

1.18 

Length 

Area 

Density 

AT 

Density 

AT 

Density 

AT 

Yoke 

12.00 

95.0 

33,600 

600 

37,300 

780 

41,000 

1,000 

Pole       

11.00 

78.5 

81,000 

280 

90,000 

528 

99,000 

1,000 

Air  gap  

0.22 

109.0 

1.13X49,500 

3,840 

1.13X55,000 

4,270 

1.13X60,500 

4,700 

Tooth  (min.)  .  . 

1.45 

39.5 

136,800 

152,000 

167,000 

133,000 

830 

147,000 

1,550 

160,000 

2,000 

Armature  core. 

6.00 

36.7 

73,800 

80 

82,000 

108 

90,200 

200 

Field-ampere  turns  per 

pole    ' 

5,630 

...          j  7,236 

8,900 

I 

^ 

cxio6 

0    h         a/ 

f 

01! 

/ 

A 

*i 

ffl 

-1 

S 

/ 

'  3 

/ 

<^>-f£   > 
c    =3400 

b      =P840° 

c 

250 
225 
200 
175 
„  160 

l» 

>  100 
75 
50 
25 


TO! 


Field  Ampere  Turns  per  Pole  foi 
Teeth  and  Armature 


>  Gap 


Field  Ampere  Turns  per  Pole 

FIG.  256. — Saturation  curves. 


FIG.  255. — Cross-magnetizing  effect. 

228.  Field  Winding. — The  field  coils  must  be  designed  with 
the  required  numbers  of  ampere-turns  and  with  sufficient  surface 
to  radiate  the  heat  due  to  the  field  copper  loss. 

If  the  coils  are  made  thick  it  will  be  difficult  to  find  space  for 
them  between  the  poles  and  they  will  be  liable  to  get  hot  in  the 
center  layers.  A  good  average  thickness  is  2  in.;  assuming  a 
current  density  in  the  field  copper  of  about  800  amp.  per  square 
inch  or  1,500  cir.  mils  per  ampere  and  a  space  factor  of  0.6  foi 
the  winding,  it  is  possible  to  place  800  X  0.6  X  2  =  approximate1^ 
1,000  ampere-turns  on  each  inch  of  the  coil  space. 

The  radiating  surface  of  the  field  coil  is  taken  as  the  external 
cylindrical  surface;  this  neglects  the  surface  of  the  ends  of  the 
coil  and  the  internal  surface  from  which  heat  is  transferred  to  the 
pole.  The  efficiency  of  the  external  radiating  surface  depends  on 
the  speed  of  the  armature  and  its  consequent  fanning  action. 
About  0.6  watts  per  square  inch  can  be  radiated  from  an  ordinary 
field  coil  for  a  temperature  rise  of  40°C.  For  very  short  thick 

18 


274  ELECTRICAL  ENGINEERING 

coils  a  higher  value  may  be  used  since  the  neglected  surface  of 
the  ends  is  so  large. 

229.  Current  Density  in  Field  Windings.— The  current  den- 
sities employed  in  shunt-field  windings  vary  from  500  amp.  per 
square  inch  for  small  slow-speed  machines  to  1,000  amp.  per 
square  inch  for  large  high-speed  machines  with  ventilated  field 
coils. 

230.  Size  of  Wire  for  the  Field  Winding.— Let  Ef  =  voltage 
across  one  coil,  //  =  current,  T/  =  number  of  turns,  lm  =  mean 
turn  in  inches,  A  =  section  of  wire  in  circular  mils  and  Rf  — 
resistance  of  the  coil. 

_  v   T 

19  '         7    T 

At  60°C.  Rf  =  12  X  --J--  =   mjL 
and 

Ef  =  IfRf  =  IfTf  -j^- 

or,  the  section  of  the  wire  is 

A  =  i&  X  lm;  (276) 

tif 

therefore  the  size  of  the  wire  to  be  used  is  fixed  when  the  ampere- 
turns  and  the  voltage  per  coil  are  known. 
Design  of  the  field  winding. 

Voltage  taken  up  in  the  field  rheostat  =  20  per  cent,  of  250  =  50  volts. 

250 5Q 

Voltage  across  one  field  coil  =  Ef  =  —  -  =  25  volts 

o 

Ampere-turns  per  coil  —  I/Tf  =  7,236. 
Thickness  of  coil  =  2  in.  (assumed). 
Mean  turn  =  38.3  in. 

7  236 
Section  of  the  wire  =  A  =  -~-  X  38.3  =  11,200  circ.  mils 

This  is  between  No.  9  and  No.  10.  B.  &  S.  wire  (Art.  86). 

Either  use  an  intermediate  size  or  wind  half  of*each  coil  with  No.  9  and 

half  with  No.  10. 

Diameter  of  No.  9    d.c.c.  =  0.126. 

Diameter  of  No.  10  d.c.c.  =  0.114. 

Average  =  0.12. 

Winding  space  =  7.2  X  2  sq.  in. 

72 
Turns  per  layer  =  Q-^  =  60. 

2  « 

Layers  =  ^^  =  16- 

Turns  per  coil  =  Tf  =  60  X  16  =  960 
Current  =  //  =  --  =  7.5  amp. 


DESIGN  OF  A  DIRECT-CURRENT  GENERATOR    275 

Power  lost  =  25  X  7.5  =  188  watts. 

Radiating  surface  =  7.2  X  3.14  X  14.2  =  320  sq.'in. 

1  oo 

Watts  per  square  inch  =  ^Q  =  ^-59. 
Temperature  rise  =  40°C.  (Art.  228). 

231.  Resistance  of  the  Armature  Winding.  —  The  length  of  a 
single  armature  conductor  is  given  approximately  by  the  empiri- 
cal formula, 

I  =  1.35r  +  Le  +  3  in.  (277) 

If  the  section  of  the  conductor  is  A  circ.  mils,  its  resistance  at 
60°C.  is  approximately          «..; 

length  in  feet  .  rt  12        /     , 

rc  =  P  -  -^-—.  -  :  ---  T-  =  12  -r  =  -,-•  ohms. 
section  in  circ.  mils  A       A 

g 

The  number  of  conductors  in  series  between  brushes  is  —  and  the 

Pi 

number  of  paths  in  multiple  is  p\  and   therefore,  the  armature 

resistance  is 

Z 

rc— 


r    -  _.    _        r 
°~     Pi      "  Pi'  ' 


(278) 


624  „  1.35  X  14.13  +  11+3 

Armature  resistance  =  ra  =  -55-  X  -  a,n  CAn  -  =  0.0052  ohms. 

8^  ol7,500 

Armature  resistance  drop  at  full  load  Ira  =  1,000  X  0.0052  =  5.2  volts. 
Armature  copper  loss  =  IJra  =  1,0002  X  0.0052  =  5,200  watts. 

232.  Determination  of  the  Magnetomotive  Force  of  the  Series 
Winding.  —  In  the  case  of  a  flat-compound  generator  the  series 
excitation  must  be  great  enough  to  counteract  the  demagnetizing 
and  cross-magnetizing  m.m.f.  of  the  armature  and  the  armature 
resistance  drop. 

Demagnetizing  Armature  Magnetomotive  Force.  —  If  the  brushes 

are  advanced  10  per  cent,  of  the  pole  pitch  the  demagnetizing 

71  78  000 

ampere-turns    per    pole  =  0.2    -~  =  0.2    <rh~¥  =  972.     The 

&p  &  X  o 

series-field  m.m.f.  required  to  overcome  this  armature  m.m.f. 
=  972  X  full-load  leakage  factor  =  972  X  1.25  =  972  +  243  = 
1,215  ampere-turns  per  pole. 

Cross-magnetizing  Armature  Magnetomotive  Force.  —  The  effect 
of  the  cross-magnetizing  armature  ampere-turns  is  difficult  to 
determine  accurately  but  some  approximation  to  it  can  be  made 
as  follows.  The  cross-magnetizing  ampere-turns  per  pole  = 


276  ELECTRICAL  ENGINEERING 

ZI  78  000 

0.8  X  -^  =  0.8  X  y^-g  =  3,888.     The  turns  beyond  the  pole 

tips  act  on  a  path  of  large  reluctance  and  may  be  neglected, 
while  the  remaining  ampere-turns  directly  under  the  poles 

71  78  000 

=  ^  Tj-5  =  0.7  X  2x8   =  ^'40°  magnetize  at  one  pole  tip  and 

increase  the  flux  and  demagnetize  at  the  other  pole  tip  and  de- 
crease the  flux.  The  magnetic  circuit  on  which  they  act  is  that 
through  the  armature  and  teeth  and  across  the  air  gap,  but  not 
through  the  pole  and  yoke. 

Fig.  255  shows  the  field  ampere-turns  per  pole  required  for  the 
armature,  the  teeth  and  the  gap  for  various  values  of  flux  in  the 
gap;  ob  is  the  field  m.m.f.  required  to  produce  normal  flux  across 

ZI 

the  gap  and  through  the  teeth  and  armature;  be  =  \f/  -~-^  = 

zp 

3,400  is  the  demagnetizing  m.m.f.  at  one  pole  tip  and  the  flux 
there  is  decreased  from  ch  to  cj;  be  =  be  =  3,400  is  the  magnetiz- 
ing m.m.f.  at  the  other  pole  tip  and  the  flux  there  is  increased 
to  el.  At  the  center  of  the  pole  the  m.m.f.  and  the  flux  are  not 
changed.  The  total  flux  o<f>  in  the  gap  at  no  load  may  be  taken 
as  proportional  to  chfe;  due  to  cross-magnetizing  a  flux  propor- 
tional to  area  hgj  is  lost  and  a  flux  proportional  to  area  glf  is 
gained;  the  result  is  a  loss  of  flux  and  voltage  proportional  to 
hgj  —  glf.  The  flux  in  the  gap  is  now  ofa  where  fa  is  found  by 
making  the  area  hjiffi  =  area  hgj  —  area  glf.  The  length  kg\ 
represents  the  effect  of  the  cross-magnetizing  armature  m.m.f. 
as  a  number  of  demagnetizing  ampere-turns  =  600  ampere- 
turns  in  this  case.  The  series  field  m.m.f.  required  to  overcome 
cross-magnetizing  =  600  X  1.25  =  750  ampere-turns  per  pole. 

The  loss  of  flux  due  to  oross-magnetizing  does  not  increase 
directly  as  the  current  but  at  a  faster  rate;  it  also  depends  very 
materially  on  the  saturation  of  the  magnetic  circuit  especially 
the  teeth;  with  low  flux  densities  the  flux  added  on  one  side  is 
approximately  equal  to  the  flux  subtracted  on  the  other  side  and 
the  cross-magnetizing  effect  is  very  small. 

To  obtain  accurate  results  it  is  necessary  to  take  account  of 
the  increase  of  the  leakage  factor  under  load  due  to  the  presence 
of  the  armature  m.m.f.  (Art.  219).  In  this  case  the  increase  is 
from  the  assumed  no-load  value  1.18  to  1.25.  The  flux  densities 
in  the  pole  and  yoke  are  increased  in  the  same  proportion  and  an 
increase  of  170  ampere-turns  per  pole  is  required  on  the  field  on 
this  account. 


DESIGN  OF  A  DIRECT-CURRENT  GENERATOR    277 

In  constructing  the  full-load  saturation  curve  of  a  genera- 
tor as  in  Art.  147,  it  is  convenient  to  treat  the  demagnetizing 
ampere-turns  as  a  constant  and  to  assume  that  they  can  be  over- 
come by  an  equal  number  of  series-field  ampere-turns;  the  extra 
turns  required  on  account  of  the  leakage  factor  =  0.25  X  972  = 
243  may  be  included  under  cross-magnetizing  as  also  the  extra 
turns  for  the  pole  and  yoke.  With  this  understanding  the  series- 
field  turns  required  to  overcome  armature  demagnetizing  = 
972  and  to  overcome  armature  cross-magnetizing  =  750  +  243 
+  170  =  1,163.  The  cross-magnetizing  armature  ampere-turns 
per  pole  =  3,888  and  the  series-field  turns  required  to  overcome 
them  =  1,163  =  approximately  30  per  cent,  of  the  cross-magnet- 
izing ampere-turns  per  pole.  For  points  higher  up  on  the  satura- 
tion curve  a  larger  percentage  than  30  would  be  required.  The 
leakage  factor  decreases  with  a  decrease  of  field  m.m.f.  and  so  the 
cross-magnetizing  effect  may  still  be  assumed  to  be  negligible 
at  the  lower  points  on  the  full-load  saturation  curve. 

Fig.  256  shows  the  no-load  and  full-load  saturation  curves  for 
the  250-kw.,  250-  volt  generator;  oa  =  7,236  =  the  field  ampere- 
turns  per  pole  required  at  no  load  to  produce  the  rated  voltage 
250  volts;  mp.  =  the  field  m.m.f.  required  at  full  load  to  over- 
come armature  reaction  =  972  +  1,163  =  2,135;  pq  =  the  vol- 
tage consumed  in  the  resistance  of  the  armature  and  series  field 
=  8  volts;  q  is  therefore  a  point  on  the  full-load  saturation  curve. 
The  other  points  may  be  found  as  explained  in  Art.  147.  The 
series-field  ampere-turns  required  to  overcome  the  resistance 
drop  =  pn  =  865  and,  therefore,  the  ampere-turns  per  pole  re- 
quired in  the  series-field  winding  for  flat-compounding  =  mn  = 
ad  =  2,135  +  865  =  3,000. 

233.  Design  of  the  Series-Field  Winding.— 

The  series-field  ampere-turns  per  pole  =  3,000. 
The  load  current  =  1,000  amp. 

Turns  per  series  coil  =  j2^  0  =  3' 

Use  3^  turns  per  coil  and  shunt  the  part  of  the  current  not  required, 
through  a  diverter. 


The  current  in  the  winding  =    '  1     =  860  amperes. 

Current  density  in  the  series  winding  =  1,000  amp.  per  square  inch 
(assumed). 


Size  of  conductor  =  .,  nnn  =  0.86  sq.  in. 

1,UUU 


278  ELECTRICAL  ENGINEERING 

Use  three  strips  2.5  X  0.125  in.  connected  in  parallel. 
Actual  section  of  conductor  =  3  X  2.5  X  0.125  =  0.94  sq.  in. 
Length  of  the  mean  turn  =  36  in.  =3  ft. 

Resistance  of  one  coil  at  60°C.  =  12  X  --  ^~r—  =  1-03  X  10~4  ohms. 

0.94  X  -106 

7T 

Voltage  drop  in  one  coil  =  860  X  1.03  X  10~4  =  0.088  volts. 

Power  lost  =  0.088  X  860  =  75  watts. 

The  cylindrical  radiating  surface  of  the  coil  =  2.5  X  40  =  100  sq.  in. 

Watts  per  square  inch  of  surface  =  y^  =  0.75,  which  is  quite  satisfactory. 

The  voltage  drop  in  the  series-field  winding  =  8  X  0.088  =  0.7  volts. 
Power  lost  in  the  series-field  winding  =  8  X  75  =  600  watts. 
234.  Design  of  the  Commutator.  — 
Winding  is  one  turn  multiple,  short  pitch. 

7          AQ  A. 

Numbers  of  commutator  segments  =  2  =  ~W  =312. 
Diameter  =  0.75Da  =  0.75  X  36  =  27  in. 

o    1  A     vy1    07 

Width  of  segment  -f  mica  =        ~^~  —  "   =  0.271  in. 

Width  of  mica  =  ^2  in.  =  0.031  in. 
Width  of  segment  =  0.24  in. 

Brush  arc  should  not  cover  more  than  three  segments  =  3  X  0.271  = 
0.813  in. 

Take  brush  arc  =  0.75  in.  =  %  in. 
Sets  of  positive  brushes  =  4. 

1  000 
Current  per  set  =    '         =  250  amp. 

Current  density  in  the  brush  =  35  amp.  per  square  inch. 

250 

Section  of  brushes  per  set  =  -^-  =  7.5  sq.  in. 

oo 

Length  of  brushes  per  set  =       r=  =  10  in. 


Brushes  per  set  =  5  —  (%  in.  X  2  in.). 
Length  of  commutator  =  12  in. 

27 
Peripheral  velocity  =  3.14  X  ^  X  400  =  2,825  ft.  per  second. 

Drop  of  voltage  at  each  brush  contact  =  1  volt. 

Loss  of  power  at  the  brush  contacts  =  1,000  X  1  X  2  =  2,000  watts. 

Brush  pressure  =  2.5  Ib.  per  square  inch. 

Total  brush  pressure  =  2.5  X  7.5  X  8  =  150  pounds. 

Coef  .  of  friction  =  0.3. 

Brush-friction  loss  =  150  X  0.3  X  2,825  =  127,000  ft.-lb.   per  minute. 


Radiating  surface  of  commutator  =  cylindrical  surface  =  3.14  X  27  X 
12  =  1,020  sq.  in. 
Total  watts  lost  =  2,000  +  2,900  =  4,900  watt. 

4  900 
Watts  per  square  inch  =    '  ^Q  —  4.8. 

Temperature  rise  will  be  under  40°C.     (Fig.  247). 


DESIGN  OF  A  DIRECT-CURRENT  GENERATOR    279 


235.  Losses  and  Efficiency. — The  core  losses  cannot  be  calculated  ac- 
curately by  any  formula  but  fairly  satisfactory  values  may  be  obtained 
by  the  use  of  curves  such  as  those  in  Fig.  257  compiled  from  the  results 
of  tests  on  completed  machines.  The  loss  in  the  teeth  must  be  cal- 
culated separately  from  that  in  the  core  since  the  densities  are  different. 

Cycles  per  Second 
15  20  25  30     40         50          00  80  100 


0         5         10         15        20        25         30        35        40        45       50 
"Watts  per  Lb. 

FIG.  257.  —  Iron-loss  curves  for  direct-  current  machines. 
Weight  of  iron  in  the  teeth 

-  156  X  0.35  X  1.45     X  9.05  X  0.28  =  200  Ib. 


=  |  ( 


where  0.28  is  the  weight  of  a  cubic  inch  of  iron. 

400      8 
The  frequency  is  -^  X  7,  =  27  cycles  per  second. 

Watts  lost  at  140,000  lines  per  square  inch  =  10  watts  per  pound. 
Watts  lost  in  the  teeth  =  200  X  10  =  2,000  watts. 

Ar  X  33A2      TT  X  252\ 
Weight  of  iron  in  the  core  below  the  teeth  =  (  -  -.  -  --  r  —  )  X 

9.05  X  0.28  =  940  Ib. 

Watts  lost  per  pound  at  density  of  80,600  lines  per  square  inch  =  3.3. 

Loss  =  940  X  3.3  =  3,100  watts. 

Total  core  loss  =  2,000  +  3,100  =  5,100  watts. 

Shunt-field  copper  loss  =  EIf  =  250  X  7.5  =  1,875  watts. 

Series-field  copper  loss  =     600  watts. 

Armature  copper  loss  =  5,200  watts. 

Brush-contact  loss  =  2,000  watts. 

Brush-friction  loss  =  2,900  watts. 

Windage  and  journal  friction  loss  =  1  per  cent.  =  2,500  watts  (assumed). 
Total  losses  =  20,175  watts. 

Efficiency  at  full  load  =  output  Bosses  10°  Per  Cent'  = 

250  000 
250,000  +  20,175  X  10°  per  Cent'  =  92'6  per  Cent' 


CHAPTER  IX 
SYNCHRONOUS  MACHINERY 

236.  Alternator. — An  alternating-current  generator  or  alter- 
nator in  its  simplest  form  consists  of  an  open  coil  of  wire  revolving 
at  uniform  speed  in  the  magnetic  field  between  a  pair  of  unlike 
poles  (Fig.  258). 

The  fields  are  excited  by  direct  current  from  a  separate  machine 
called  an  exciter  at  125  or  250  volts. 


FIG.  258. — Single-phase  alternator,  revolving-armature  type. 

Between  the  slip  rings  a  and  b  an  alternating  e.m.f .  is  generated 
of  instantaneous  value 


=  n  ~  10~8  volts 
at 


(279) 


where  n  is  the  number  of  turns  in  the  coil  and  -77  is  the  rate  of 

change  of  the  flux  interlinking  with  the  coil  or  the  rate  at  which  the 
coil  is  cutting  the  flux.  The  result  is  the  same  if  the  armature  is 
stationary  and  the  field  revolves. 

237.  Types  of  Alternators. — There  are  three  principal  types 
of  alternators: 

(a)  Revolving  armature. 

(b)  Revolving  field. 

(c)  Inductor. 

280 


SYNCHRONOUS  MACHINERY 


281 


Type  (a)  is  illustrated  in  Fig.  258.  The  field  poles  are  sta- 
tionary and  the  armature  revolves  between  them.  The  ends  of 
the  winding  are  brought  out  to  two  slip  rings  in  single-phase 
machines  and  to  three  or  more  slip  rings  in  polyphase  machines 
and  the  current  is  collected  by  copper  or  carbon  brushes. 

The  armature  is  necessarily  of  small  size  since  the  peripheral 
speed  is  limited  and  there  is  very  little  space  for  insulation.  The 
armature  conductors  are  also  acted  upon  by  centrifugal  forces 
which  tend  to  throw  them  out  of  the  slots.  The  revolving  arma- 
ture is  therefore  only  suitable  for  machines  of  small  size  and  low 
voltage.  It  is,  however,  necessary  in  the  case  of  rotary  converters 
where  the  same  armature  winding  carries  both  alternating  and 
direct  currents. 

(6)  The  revolving  field  type  illustrated  in  Fig.  259  and  Fig. 
260,  is  almost  universal  for  all  sizes  and  voltages.  The  armature 
is  the  stationary  part  and  the  field  poles  revolve.  This  type  has 
many  advantages  over  the  revolving  armature  type.  (1)  It 


FIG.  259. — Single-phase  alternator,  revolving-field  type. 

requires  only  two  slip  rings  even  for  polyphase  machines  and  these 
slip  rings  carry  only  the  direct  current  supplied  to  the  field  wind- 
ing, while  the  load  current  is  taken  off  from  stationary  terminals. 
(2)  There  is  much  more  space  for  the  armature  windings  and 
they  are  relieved  from  all  centrifugal  strains.  They  can,  there- 
fore, be  much  better  insulated  and  ventilated.  The  field  wind- 
ings are  made  of  copper  strap  and  the  revolving  member  is  very 
rugged  and  is  not  affected  by  strains  due  to  rotation;  thus,  much 
higher  peripheral  speeds  may  be  used  than  with  type  (a),  with 
consequent  increase  in  economy  of  material. 

The  revolving-field  members  are  made  in  two  forms,  the 
salient  pole  rotor  and  the  cylindrical  rotor.  The  rotor  with 
salient  poles,  Figs.  259  and  260,  is  used  almost  universally  for 
machines  of  all  but  the  largest  outputs  and  highest  speeds. 


282 


ELECTRICAL  ENGINEERING 


Cylindrical  rotors,  Fig.  335,  are  employed  for  turbo-alternators 
of  very  large  output  which  must  run  at  extremely  high  speeds  in 
order  to  obtain  the  maximum  output  per  pound  of  material. 
The  peripheral  speeds  run  up  to  25,000  ft.  per  minute  and  very 
large  stresses  are  developed  due  to  centrifugal  force.  The  rotors 
are  made  of  solid  steel  castings  or  of  a  number  of  steel  discs 
bolted  together  and  have  slots  milled  out  to  carry  the  field  wind- 
ings. The  shaft  is  bolted  or  dovetailed  to  the  ends  of  the  rotor. 


Windln, 

Blip  Rings  for  Excitation 

iDBUla 


Field.8plder 


FIG.  260. — Revolving-field   alternator. 

The  cylindrical  rotor  gives  an  air  gap  of  practically  uniform 
length  and  reluctance  over  the  whole  periphery  and  the  field 
m.m.f.  is  distributed.  It  lends  itself  more  readily  to  analytical 
treatment  than  the  salient  pole  rotor  with  its  varying  gap  re- 
luctance and  concentrated  field  m.m.f.  The  vector  diagrams  in 
Figs.  303  to  306  give  very  satisfactory  results  for  machines  with 
cylindrical  rotors  but  one  more  approximate  for  those  with  salient 
poles. 

(c)  The  inductor  alternator  is  almost  obsolete.  In  these 
machines  the  field  and  armature  windings  are  both  stationary 
and  a  part  of  the  iron  of  the  magnetic  circuit  revolves,  producing 
a  periodic  pulsation  of  the  reluctance  of  the  magnetic  circuit  and 
consequently  a  variation  of  the  flux  linking  with  the  armature 
winding.  Fig.  261  represents  one  type  of  inductor  alternator: 


SYNCHRONOUS  MACHINERY 


283 


ff  is  the  stationary  field  winding  which  produces  the  magnetic 
flux  <£,<£,  in  the  direction  indicated.  a,b,  a,6,  are  the  armature 
coils  which  may  be  connected  either  in  series  or  in  multiple.  / 
is  the  revolving  part  of  the  magnetic  circuit  and  is  called  the  in- 
ductor. The  polar  projections  on  it  are  all  north  poles  but  the 
amount  of  flux  issuing  from  7  and  linking  with  the  armature 
coils  a,a,  6,6,  depends  on  the  relative  position  of  the  inductor 
projections  and  the  projections  carrying  the  armature  coils. 

In  Fig.  262,  curve  1  represents  the  variation  of  the  flux  <j>a 
interlinking  with  one  armature  coil  a  starting  from  the  position 
when  this  flux  is  maximum,  Fig.  261.  Curve  2  represents  the 
e.m.f.  generated  in  coil  a  by  the  varying  flux.  As  0^  decreases 


FIG.  261. — Inductor  alternator. 


FIG.  262. — Fluxes  and  e.m.fs.  in  an 
inductor  alternator. 


an  e.m.f.  is  generated  in  the  positive  direction  and  as  it  increases 
again  an  e.m.f.  is  generated  in  the  negative  direction.  Thus  al- 
though the  flux  does  not  reverse  its  direction  and  never  reaches 
zero,  an  alternating  e.m.f.  is  generated  in  the  armature  coil. 
If  the  pole  pieces  are  properly  shaped  a  sine  wave  of  e.m.f.  will 
be  produced.  Curves  3  and  4  represent  the  variation  of  flux 
and  e.m.f.  in  coil  b.  The  e.m.f.  in  coil  6  is  displaced  180  degrees 
from  that  in  a  and  before  the  two  are  connected  in  series  the 
terminals  of  one  must  be  reversed. 

Inductor  alternators  were  very  heavy  and  expensive  and  have 
been  superseded  by  the  other  types. 

Any  one  of  these  three  types  may  be  wound  as  single-phase  or 
polyphase  machines. 

Alternators  are  divided  into  classes  depending  on  the  type  of 
prime  mover  employed:  (a)  Engine  type,  (6)  waterwheel  type, 
(c)  steam  turbine-driven  type. 


284  ELECTRICAL  ENGINEERING 

(a)  Engine-type  alternators  are  direct-connected  to  recipro- 
cating steam  engines  and  they  require  a  large  flywheel  effect  to 
ensure  uniform  angular  velocity. 

(6)  Waterwheel-type  alternators  range  in  speed  from  50 
to  400  r.p.m.  depending  largely  on  the  available  head  of  water. 
The  angular  velocity  of  water  turbines  is  uniform  and  flywheels 
are  not  required.  Water  wheel  alternators  may  be  either  hori- 
zontal or  vertical. 

(c)  Steam  turbine-driven  alternators  or  turbo-alternators  are 
extra  high-speed  generators  usually  750  or  1,500  r.p.m.  for  25 
cycles  and  1,800  or  3,600  r.p.m.  for  60  cycles,  corresponding  to 
four-pole  and  two-pole  rotors  respectively.  Cylindrical  rotors 
are  ordinarily  employed  on  account  of  their  greater  mechanical 
strength  and  smaller  windage  loss.  On  account  of  the  large 
output  per  pound  of  material  forced  ventilation  is  required  to 
keep  the  temperature  within  the  required  limits  and  therefore 
such  machines  must  be  entirely  enclosed. 

238.  Electromotive  Force  Equation.  —  Fig.  258  represents  a 
two-pole,  single-phase  alternator.  The  armature  winding  is  a 
single  coil  of  n  turns  revolving  at  a  constant  speed.  The  magnetic 
field  is  assumed  to  be  uniform. 

The  e.m.f.  generated  in  the  winding  goes  through  one  complete 
cycle  during  each  revolution,  and  thus  the  frequency  in  cycles 
per  second  is  equal  to  the  speed  in  revolutions  per  second  or  /  = 
rev.  per  sec. 

The  angular  velocity  of  the  coil  is  co  radians  per  second,  and 
therefore 

w  =  27r  X  rev.  per  sec.  =  2irf  (280) 

If  <f>  is  the  maximum  flux  inclosed  by  the  coil,  that  is,  the  flux 
inclosed  when  the  coil  is  vertical,  as  shown,  and  time  is  measured 
from  this  instant,  then  at  time  t,  after  the  coil  has  turned  through 
an  angle  6,  the  flux  inclosed  is  0  =  $  cos  0,  and  the  e.m.f.  gener- 
ated in  the  coil  is 

e=  -rAo-8 
at 


but  d  =  ut  =  2irft,  and  thus 

e  =  -  ndj(3>c 
at 

=  2vfn3>  10~8  sin  2irft  volts  (281) 

=  Em  sin  6. 


SYNCHRONOUS  MACHINERY  285 

This  is  a  sine  wave  of  maximum  value 

Em  =  27r/7i$  10~8  volts  (282) 

and  effective  value 

V 

E  =  — 5|  =  4.44/71$  10-8  volts  (283) 

This  is  the  e.m.f.  equation  for  an  alternator  which  produces  a 
sine  wave  of  e.m.f.  and  has  a  concentrated  winding,  that  is,  all 
the  turns  wound  in  a  single  coil. 

This  result  may  also  be  obtained  as  follows.  The  flux  cut  per 
second  by  each  turn  of  the  coil  is  4/$  lines,  and  therefore  the 
average  e.m.f.  generated  in  the  coil  is 

Eavg  =  4/n$  10-8  volts,  (284) 

but  for  a  sine  wave  the  ratio  of  the  maximum  to  the  average 
ordinate  is  ~'  and  therefore  the  maximum  e.m.f.  is 

Em  =  I  Eaug  =  27r/7i$  10~8, 
and  the  effective  value  is  as  before 

Tjl 

E  =  — ^  =  4.44/7i$  10-8. 

The  e.m.f.  generated  in  an  alternator  is  directly  proportional 
to  the  frequency  /,  to  the  number  of  turns  in  series  n  and  to  the 
flux  under  each  pole  $. 

In  a  two-pole  machine  one  revolution  or  360  mechanical  de- 
grees corresponds  to  one  cycle  or  360  electrical  degrees  and  the 
frequency  is  equal  to  the  number  of  revolutions  per  second ;  in  a 
p-pole  machine  the  e.m.f.  goes  through  a  complete  cycle  when 

fY\ 

the  coil  moves  across  a  pair  of  poles  and  thus  through  ~  cycles 
in  one  revolution.     In  this  case  360  mechanical  degrees  =  |  X 

IY\ 

360  electrical  degrees  or  one  mechanical  degree  =  §  electrical 

2i 

degrees. 

The  frequency  in  a  p-pole  alternator  in  cycles  per  second  is 

/  =      X  rev.  per  sec.  (285) 


286 


ELECTRICAL  ENGINEERING 


239.  Form  Factor.  —  If  the  flux  in  the  air  gap  is  not  so  distrib- 
uted as  to  give  a  sine  wave  of  e.m.f.,  the  average  value  of  the 
generated  e.m.f.  is  still  given  by  equation  (284), 


Eavg 


10- 


but  the  ratio  of  maximum  to  average  value  is  not  ~  and  the  ratio 

of  effective  to  maximum  value  is  not  ~~7s' 

The  effective  value  of  the  general  alternating  wave  can  be 
expressed  as 

E  =  4yfn$  10~8  volts,  (286) 

where  7  is  called  the  form  factor  of  the  wave  and  is  defined  as 
the  ratio  of  the  effective  value  to  the  average  value  of  the  ordi- 
nate  of  the  wave. 

The  form  factor  of  a  rectangular  wave  is  1.00  and  for  all  other 
waves  is  greater  than  1.00.     For  a  sine  wave  it  is 

Em 


7  = 


1.11. 


240.  Polyphase  Alternating-current  Generators. — If  the  arma- 
ture of  an  alternator  carries  two  similar  windings  displaced  90 
electrical  degrees  from  one  another,  Fig.  263,  the  winding  is 

A  BCD 


.    *2  A     A    o 

FIG.  263. — Two-phase  alternator. 

a  two-phase  winding  and  the  alternator  a  two-phase  alternator. 
Two  e.m.fs.  are  produced  equal  in  value  but  displaced  90  degrees 
in  phase. 

If  the  armature  carries  three  similar  windings  displaced  from 
one  another  by  120  electrical  degrees  (Fig.  264),  the  winding  is  a 
three-phase  winding  and  the  alternator  a  three-phase  alternator. 
Three  e.m.fs.  are  produced  equal  in  value  but  displaced  120 
degrees  in  phase. 


SYNCHRONOUS  MACHINERY 


287 


If  the  windings  of  the  three  phases  start  at  «i,  s%  and  83  and  end 
at  /i,  /2  and  /s,  the  phases  may  be  interconnected  in  two  ways : 
(1)  join /i  to  s2,/2  to  s3  and/s  to  Si,  this  is  the  "  delta"  connection, 
Fig.  265;  (2)  join  si,  s2  and  s3  together  and  /i,  /2  and  /3  to  the 
three  terminals,  this  is  the  "star"  or  Y  connection  (Fig.  266). 

For  a  given  number  of  turns  per  phase  the  Y  connection  gives 
a  higher  terminal  voltage  than  the  delta  connection  and  a  corre- 
spondingly smaller  current  output. 


Rotation 


A  >/i    "•/. 


FIG.  264. — Three-phase  alternator. 


'/. 


FIG.  265. — Delta  connection.          FIG.  266. — Star  or  "Y"  connection. 


, 

A 

/I 

/, 

Si 
Jfo 

»1 

i 

Four  Pole,  Single  Phase 
Chain  Winding 

*2 
ur  Pole,  Two  Phase, 
Chaiu  Winding 

•i 

our  1 

«3                   l/i    1 
'ole,  Three  Phase, 

Chain  Winding 
FIG.  267. — Concentrated  chain  windings. 

The  e.m.fs.  currents  and  power  in  polyphase  circuits  are  dis- 
cussed in  Art.  124. 

The  electrical  power  developed  at  any  instant  in  a  single  phase 
alternator  is  the  product  of  the  instantaneous  values  of  the  e.m.f . 
and  current  (Fig.  89).  It  pulsates  between  a  maximum  positive 

TJI       T  rr       T 

value     "X  m  (1  +  cos  <£)  and  a  negative  value     "       (1  —  cos  0) 

and  its  average  value  is  El  cos  <j>. 

The  power  developed  in  a  three  phase  alternator  is  the  sum  of 
the  instantaneous  powers  developed  in  the  three  phases ;  its  value 


288 


ELECTRICAL  ENGINEERING 


at  any  instant  is  z\i\  +  eziz  +  etfz  =  3EI  cos  $  and  remains 
constant. 

The  armature  m.m.f.  of  the  single-phase  alternator  pulsates 
likewise  between  nlm  and  zero  where  n  is  the  number  of  turns  on 
the  armature,  but  it  remains  fixed  in  direction  relative  to  the 
armature  and  revolves  at  synchronous  speed  relative  to  the  poles. 


s  f 

Four  Pole,  Single  Phase,  Double  Layer  Winding 


Four  Pole,  Two  Phase,  Double  Layer  Winding 


Four  Pole.    Three    Phase.  Double  Layer  Winding 
FIG.  268. — Concentrated  double-layer  windings. 

The  armature  m.m.fs.  of  the  various  phases  of  a  polyphase 
alternator  combine  to  produce  a  constant  m.m.f.  of  armature 
reaction  fixed  in  direction  relative  to  the  poles  and  revolving  at 
synchronous  speed  relative  to  the  armature. 

The  conditions  of  operation  of  a  polyphase  alternator  or  syn- 
chronous motor  are  thus  much  more  satisfactory  than  those  of 


SYNCHRONOUS  MACHINERY 


289 


the  single-phase  machines.  Single-phase  machines  are  built 
only  in  the  smaller  sizes  except  in  special  cases  as  for  instance  to 
supply  power  to  a  single-phase  electric-railway  system. 

241.  Alternator  Windings. — There  are  a  great  many  special 
alternator  windings  but  the  majority  of  them  come  under  the  two 
classes  of  chain  windings  and  double-layer  windings. 


\     ( 


\     C 


s  if 

FIG.  269. — Four-pole,    single-phase,    chain    winding    distributed    in    six    slots 

per  pole. 

Fig.  267  shows  four-pole  single-,  two-  and  three-phase  chain 
windings  for  armatures  with  one  slot  per  phase  per  pole.  These 
windings  are  all  concentrated  windings. 

Fig.  268  shows  the  corresponding  double-layer  windings. 

Fig.  269  shows  a  four-pole  single-phase  chain  winding  distrib- 
uted in  six  slots  per  pole.  Fig.  270  shows  a  winding  for  the  same 
machine  using  only  four  of  the  six  slots  per  pole  as  explained  in 
Art.  242. 


FIG.  270. — Four-pole,   single-phase   chain  winding  using  only  four  of  the  six 

slots  per  pole. 

Fig.  271  shows  a  two-phase  chain  winding  for  the  armature  in 
Fig.  289.  The  windings  are  distributed  in  three  slots  per  phase 
per  pole. 

Fig.  272  shows  a  three-phase  chain  winding  for  the  same  arma- 
ture, distributed  in  two  slots  per  phase  per  pole. 

19 


290 


ELECTRICAL  ENGINEERING 


Figs.  273  tp  276  show  the  double-layer  windings  corresponding 
to  Figs.  269  to  272. 

242.  Distribution  Factors.— The  windings  in  Figs.  267  and 
268  are  all  concentrated  windings,  that  is,  they  are  placed  in  one 
slot  per  phase  per  pole. 


«.  w 

FIG.  271. — Four-pole,  two-phase,  chain  winding  distributed  in  three  slots  per 

phase   per   pole. 


When  a  winding  is  made  up  of  a  number  of  coils  placed  in  sepa- 
rate slots  the  e.m.fs.  generated  in  the  various  coils  are  displaced 
in  phase  and  the  terminal  e.m.f.  is  less  than  if  the  winding  had 
been  concentrated.  The  factor  by  which  the  e.m.f.  of  a  concen- 
trated winding  must  be  multiplied  to.  give  the  e.m.f.  of  a  distrib- 


J/2  1/3 

FIG.  272. — Four-pole,  three-phase,  chain  winding  distributed  in  two  slots  per 

phase  per  pole. 


uted  winding  of  the  same  number  of  turns  is  called  the  distribu- 
tion factor  for  the  winding  and  it  is  always  less  than  unity. 

When  a  single-phase  winding  is  distributed  in  two  slots  per 
pole  spaced  at  90  degrees  the  e.m.fs.  in  the  two  coils  are  90  de- 
grees out  of  phase.  If  the  effective  value  of  the  e.m.f.  generated 


SYNCHRONOUS  MACHINERY 


291 


FIG.  273. — Four-pole,  single-phase,  double-layer  winding  distributed  in  six  slots 

per  pole. 


FIG.  274. — Four-pole,  single-phase,  double-layer  winding  using  only  four  of  the 

six   slots   per  pole. 


FIG.  275. — Four-pole,  two-phase,  double-layer  winding  distributed  in  three  slots 
per  phase  per  pole. 


FIG.  276. — Four  pole,  three-phase,  double-layer  winding  distributed  in  two  slots 
per  phase  per  pole. 


292 


ELECTRICAL  ENGINEERING 


in  each  coil  is  e,  then  the  terminal  e.m.f.  is  et  —  -\/2e,  Fig.  277, 
and  the  distribution  factor  is 


8  =    L  = 
2e 


2e 


0.70/7 


When  a  single-phase  winding  is  distributed  in  three  slots  per  pole 
spaced  at  60  degrees  the  terminal  e.m.f.  is  the  sum  of  three  e.m.fs. 
e  at  60  degrees  to  one  another.  It  is  et  =  2e}  Fig.  278,  and  the 
distribution  factor  is 


=  0.666. 


0.705 


FIG.  278. 


When  a  single-phase  winding  is  distributed  in  six  or  more  slots 
per  pole  the  distribution  factor  may  be  taken  as 


8  =       =  o.64. 

7T 

In  Fig.  279  the  semi-circumference  represents  the  e.m.f.  of  the 
concentrated  winding  and  the  diameter  represents  the  e.m.f.  of 
the  distributed  winding. 

The  e.m.fs.  in  the  coils  from  b-c  add  very  little  to  the  terminal 
e.m.f.  and  this  part  of  the  winding  is  usually  omitted  and  the 

terminal  e.m.f.  is  decreased  in  the  ratio  —  =  cos  30°  =  0.866,  or 

oc 

is  decreased  13.4  per  cent,  while  the  resistance  and  reactance  of 
the  winding  are  decreased  33  J^  per  cent. 
The  distribution  factor  for  this  case  is 

g  _  06  _  3|le 
4e         4:6 

Figs.  270  and  274  show  four-pole  single-phase  windings  with 
only  four  of  the  six  slots  per  pole  used. 


SYNCHRONOUS  MACHINERY 


293 


The  terminal  e.m.f.  of  a  two-phase  winding  distributed  in  two 
slots  per  phase  per  pole  is  made  up  of  two  e.m.fs.  of  value  e 
displaced  45  degrees  from  one  another.  It  is  et  =  1.848  e,  Fig. 
280,  and  the  distribution  factor  is 

et         1.848e 


2e 


0.924. 


FIG.  279. 


FIG.  280. 


FIG.  281. 


For  a  two-phase  winding  with  three  slots  per  phase  per  pole, 
Figs.  271  and  275,  the  terminal  e.m.f.  is  made  up  of  three  e.m.fs. 
displaced  30  degrees  from  one  another.  It  is  et  =  2.733e,  Fig. 
281,  and  the  distribution  factor  is 

2"733e  =  0.911. 


FIG.  282. 


FIG.  283. 


The  terminal  e.m.f.  of  a  three-phase  winding  distributed  in 
two  slots  per  phase  per  pole,  Figs.  272  and  276,  is  made  up  of  two 


294 


ELECTRICAL  ENGINEERING 


e.m.fs.  of  value  e  displaced  30  degrees  from  one  another.     It  is 
et  =  1.9316,  Fig.  282,  and  the  distribution  factor  is 

«*       1.93'le 


2e 


=  0.966. 


With  three  slots  per  phase  per  pole  the  factor  is 
6  =  0.96,  Fig.  283. 

The  following  table  gives  the  distribution  factors  for  single-, 
two-  and  three-phase  windings. 


Slots  per  phase 
per  pole 

Distribution  factor 

Single-phpse 

Two-phase 

Three-phase 

1 

1.000 

1.000 

1.000 

2 

0.705 

0.924 

0.966 

3 

0.666 

0.911 

0.960 

4 

0.653 

0.906 

0.958 

6 

0.64 

0.903 

0.956 

6 

0.83,  if  only  two-thirds  of  slots  are  used. 

243.  Multiple -circuit  Windings. — The  windings  already  dis- 
cussed are  all  single  circuit,  that  is,  all  the  turns  of  one  phase  are 
connected  in  series.     In  low- voltage  machines  with  a  large  current 
output  it  is  necessary  to  connect  the  coils  forming  each  phase  in 
multiple  circuit.     When  connected  two-circuit  the  terminal  e.m.f. 
is  reduced  to  one-half  and  the  current  output  is  doubled;  the 
power  output,  therefore,  remains  the  same. 

The  e.m.fs.  generated  in  the  sections  of  the  windings  which  are 
connected  in  multiple  must  be  of  the  same  value  and  must  be  in 
phase  or  circulating  currents  will  flow.  It  is  also  necessary  that 
the  resistances  and  reactances  of  the  sections  be  of  the  same 
value  or  one  part  of  the  winding  will  supply  more  current  than 
the  other. 

Fig.  284  shows  a  four-pole  three-phase  double-layer  winding 
with  one  slot  per  phase  per  pole  connected  Y  and  A  single 
circuit  and  two  circuit.  A  winding  may  be  connected  with  as 
many  circuits  in  multiple  as  there  are  pairs  of  poles. 

244.  Short-pitch  Windings. — The  pitch  of  a  winding  is  the 
distance  between  the  two  sides  of  one  of  the  coils  forming  the 
winding.     When  the  coil  pitch  is  equal  to  the  pole  pitch  or  the 


SYNCHRONOUS  MACHINERY 


295 


distance  between  the  centers  of  adjacent  poles,  the  winding  is  full 
pitch.  When  the  coil  pitch  is  less  than  the  pole  pitch  the  winding 
is  fractional  pitch  or  short  pitch. 

In  Fig.  286  abcdf  shows  the  distribution  of  flux  under  two 
adjacent  poles  of  an  alternator.     The  area  under  the  section  of 


Star  or  Y 


Delta 


1 


M            ' 

k       I      I      <      f     f 

??  

Si 

S2          *S3   Two  Circuit 

J 

•  f 

1 

'f.      1 

Two  Circu: 
Star  or  Y 


T?    I 


Two  Circuit 
Delta 

FIG.  284. — Four-pole,  three-phase,  multiple-circuit  windings. 

the  curve  abc  or  cdf  multiplied  by  the  length  of  the  pole  parallel 
to  the  shaft  gives  the  flux  3>  crossing  the  air  gap  under  each  pole. 
As  the  side  g  of  the  coil  cuts  across  the  flux  in  the  gap  an  e.m.f. 
is  generated  in  it  of  the  same  wave  shape  as  the  flux  distribution. 
If  gh  is  a  full-pitch  coil  the  side  h  will  occupy  a  position  under  the 
adjacent  pole  similar  to  that  of  g  and  the  e.m.fs.  generated  in  the 


290 


ELECTRICAL  ENGINEERING 


two  sides  will  be  of  the  same  value  and  wave  shape  but  displaced 
180  degrees  in  phase;  they  therefore  act  in  the  same  direction 
around  the  coil  and  add  directly  to  give  the  terminal  e.m.f.  If  e 
is  the  effective  value  of  the  e.m.f.  generated  in  one  side  of  the 
coil  the  terminal  e.m.f.  is  E  =  2e.  With  a  full-pitch  concentrated 
winding  the  wave  form  of  the  generated  e.m.f.  is  the  same  as  the 
wave  of  flux  distribution  under  the  poles. 


E.M.F.  between  Terminals 


E.M.F.  in  g  or  h 


FIG.  285. — Full-pitch  coil. 

If  the  coil  pitch  is  less  than  the  pole  pitch  by  an  angle  a,  the 
e.m.f.  wave  generated  in  the  side  h  leads  the  e.m.f.  in  g  by  an 
angle  a,  Fig.  285,  and  the  terminal  e.m.f.  is  the  vector  sum  of  two 
e.m.fs.  of  effective  value  e  displaced  in  phase  by  an  angle  a.  It  is 


E  =  2e  cos  - 


(287) 


and  is  less  than  the  e.m.f.  generated  in  the  full-pitch  winding  in 

Ot 

the  ratio  cos  ~  *  1. 


E.M.F.  between 
Terminals 


E=2e  Cos  .5 


•  FIG.  286. — Short-pitch  coil. 

Fractional-pitch  windings  are  sometimes  used  in  order  to  elimi- 
nate certain  harmonics  from  the  e.m.f.  wave  of  the  generator. 
Take  the  case  of  a  machine  with  the  wave  of  flux  distribution, 
shown  in  Fig.  287,  consisting  of  a  fundamental  and  a  fifth  har- 
monic. With  a  full-pitch  winding  the  e.m.f.  wave  would  consist 


SYNCHRONOUS  MACHINERY  297 

of  a  fundamental  and  the  prominent  fifth  harmonic.  If,  how- 
ever, the  coil  pitch  is  made  only  80  per  cent,  of  the  pole  pitch  the 
e.m.f.  in  one  side  of  the  coil  will  lead  that  in  the  other  by  36 
degrees  and  the  fifth  harmonics  in  the  two  sides  will  be  in  direct 
opposition  and  will  disappear  (Fig.  288).  The  terminal  e.m.f. 
will  consist  only  of  the  fundamental  and  it  will  be  decreased  in 
ratio  cos  18  degrees:!.  To  eliminate  an  nth  harmonic  the  coil 

pitch  must  be  either  lengthened  or  shortened  by  -  th  of  the  pole 

pitch.  Thus  the  wave  form  of  the  e.m.f.  generated  in  a  short- 
pitch  winding  is  not  the  same  as  the  wave  of  the  flux  distribution 
in  the  air  gap. 

E.M.F.between- 
Terminals 

.Curve  of  Flux  Distribution 


FIG.   287.— Flux  wave   with   fifth     FIG.  288. — Elimination  of  fifth  harmonic, 
harmonic. 

Similarly  the  wave  form  of  any  distributed  winding  differs 
from  the  wave  of  flux  distribution,  since  the  terminal  e.m.f.  is 
the  sum  of  a  number  of  waves  displaced  from  one  another.  A 
fully  distributed  winding  gives  an  e.m.f.  wave  of  approximately 
sine  form  at  no  load  regardless  of  the  flux  distribution. 

245.  Effect  of  Distributing  the  Winding. — (1)  The  core  is  used 
to  better  advantage  since  a  number  of  small  slots  evenly  spaced 
are  used  instead  of  a  few  large  ones.     (2)  The  copper  is  evenly 
distributed  over  the  armature  surface  and  thus  the  copper  loss  is 
also  distributed  and  the  heat  developed  by  it  can  more  easily  be 
dissipated.     A  higher  current  density  in  the  copper  can,  therefore, 
be  used.     (3)  The  self-inductive  reactance  is  very  largely  de- 
creased by  distributing  the  winding  in  a  large  number  of  slots, 
since  the  coefficient  of  inductance  of  a  coil  is  proportional  to  the 
square  of  the  number  of  turns.     (4)  The  terminal  e.m.f.  is  de- 
creased as  shown  in  Art.  242  but  the  wave  form  is  made  more 
nearly  sinusoidal. 

246.  Harmonics  Due  to  the  Teeth. — Fig.  289  shows  two  posi- 
tions of  the  armature  of  an  alternator  relative  to  the  field  poles. 
In  A  there  are  four  teeth  under  the  pole  and  three  slots,  while  in 


298  ELECTRICAL  ENGINEERING 

B  there  are  three  teeth  and  four  slots.  The  reluctance  of  the 
magnetic  circuit  of  the  generator  is  a  minimum  in  A  and  the 
flux  in  the  circuit  is  a  maximum,  while  in  B  the  reluctance  is  a 
maximum  and  the  flux  is  a  minimum.  If  there  are  a  teeth  or 
slots  per  pole  the  flux  per  pole  pulsates  once  in  the  distance  of  a 
slot  pitch  or  2a  times  in  two  pole  pitches.  The  frequency  of  the 
pulsations  is  2af  where  /  is  the  normal  frequency  of  the  machine. 
The  flux  per  pole  consists  of  a  constant  value  $  and  superim- 
posed on  it  an  alternating  flux  of  amplitude  $1  and  frequency  2af. 


UITUTJTLT 


m 

& 


FIG.  289. — Harmonics  due  to  the  teeth. 

Measuring  time  from  the  instant  represented  in  A  when  the 
flux  is  maximum,  the  flux  at  any  time  t  or  angle  0  is 

0P  =  $  +  $1  cos  2a0, 

and  the  flux  enclosed  by  the  coil  at  angle  0  is 
<j>p  cos  8  .=  $  cos  6  +  $1  cos  2a0  •  cos  B 

=  $  cos  B  +  ^{cos  (2a  +  1)0  +  cos  (2o  -  1)0} ; 

Zi 

the  e.m.f.  generated  in  a  coil  of  N  turns  concentrated  in  a  single 
slot  is 


e  =  -  N^-(<t>P  cos  0)  =  -  N-j-A  $  cos  0  +  ^{cos  (2o  +  1)0 
at  at/  L  z 

-f-  cos  (2a  — 

=  -  ^[-  $sin  0  +  ^{  -  (2a  +  1)  sin  (2o  +  1)0 
L  ^ 

-  (2a  -  1)  sin  (2a  - 

but 

dd  _ 

and  therefore, 

e  =  2T/ATJ$  sin  0  +  —5 — ^>i  sin  (2a  +  1)0 

+  — ^lSin  (2a-l)0| 


SYNCHRONOUS  MACHINERY        299 

a  fundamental  and  two  harmonics  the  (2a  +  l)th  and  the  (2a  — 
l)th.  With  three  slots  per  pole  the  tooth  harmonics  are  the 
seventh  and  the  fifth;  and  with  six  slots  per  pole  the  harmonics 
are  the  thirteenth  and  the  eleventh. 

The  pulsations  of  the  flux  are  opposed  by  the  large  inductance 
of  the  field  winding  and  they  are  therefore  very  small  and  in 
addition  an  nth  harmonic  current  is  opposed  by  a  reactance  in 
the  alternator  armature  and  load  circuit  which  is  n  times  as 
great  as  the  reactance  for  the  fundamental  current. 

If  the  width  of  the  pole  face  plus  a  small  allowance  for  fringing 
is  a  multiple  of  the  slot  pitch  the  variation  of  the  gap  reluctance 
is  negligible  and  the  pulsation  of  flux  does  not  occur. 

247.  Effect  of  Third  Harmonics  in  Three-phase  Alternators. — 
If  the  e.m.f .  per  phase  in  a  three-phase  delta-connected  alternator 
contains  a  third  harmonic,  triple-frequency  currents  will  circulate 
through  the  closed  delta  at  all  times.  Referring  to  Fig.  129, 
the  resultant  e.m.f.  around  the  closed  circuit  at  any  instant  is 
ei  +  ^2  +  £3  and  this  was  shown  to  be  equal  to  zero  in  the  case 
of  three  similar  sine  waves  displaced  at  120  degrees,  Art.  123. 
The  third  harmonics  are,  however,  not  combined  at  120  degrees 
but  at  3  X  120  =  360  degrees  and  are  in  phase  and  the  resultant 
e.m.f.  is  three  times  the  magnitude  of  the  third  harmonic  of  one 
phase  and  this  will  cause  a  third  harmonic  of  current  to  circulate 
through  the  closed  winding.  This  current  may  be  of  the  order 
of  full-load  current  in  the  case  of  alternators  of  low  reactance. 
The  reactance  of  the  alternator  winding  to  the  triple-frequency 
current  is  three  times  that  opposed  to  the  current  of  fundamental 
frequency  but  only  the  true  reactance  and  not  the  synchronous 
reactance  is  effective  in  limiting  the  circulating  current.  The 
third  harmonic  of  e.m.f.  does  not  appear  at  the  terminals  since 
it  is  consumed  in  producing  the  circulating  current. 

If  the  alternator  is  connected  F,  the  third  harmonic  e.m.f.  will 
not  appear  in  the  e.m.f.  between  terminals,  since  this  e.m.f.  is 
the  difference  of  two  e.m.f s.  at  120  degrees  to  one  another  or  the 
sum  of  two  e.m.fs.  at  60  degrees,  the  third  harmonics  will  be 
combined  at  3  X  60  =  180  degrees  and  will  therefore  neutralize 
one  another.  If,  however,  the  neutral  is  connected  to  ground  at 
both  the  generator  and  receiver  ends.  A  third  harmonic  of 
current  may  flow  in  the  neutral  supplied  from  the  three  phases. 

Alternators  should,  whenever  possible,  be  connected  Y  instead 
of  A  to  reduce  the  danger  of  circulating  currents. 


300  ELECTRICAL  ENGINEERING 

248.  General    Electromotive    Force    Equation. — The    e.m.f. 
equation,  derived  in  Art.  239, 

E  =  4yfn3>  10~8  volts 

which  applies  only  to  concentrated  windings  may  be  extended  to 
include  all  windings  by  introducing  the  distribution  factor  5. 

Thus  the  general  equation  for  the  effective  value  of  the  e.m.f. 
between  terminals  of  an  alternator  is 

E  =  45yfn3>  10~8  volts,  (288) 

where 

/  =  frequency  in  cycles  per  second, 

n  =  number  of  turns  in  series  between  terminals, 

3>  =  flux  from  one  pole, 

7  =  form  factor  of  the  e.m.f.  wave, 

8  =  distribution  factor  of  the  winding. 

This  equation  holds  both  for  the  single-phase  alternator  and  for 
any  phase  of  a  polyphase  alternator  with  n  turns  in  series  per 
phase. 

If  the  winding  is  short  pitch  the  e.m.f.  is  reduced  in  the  ratio 

cos  o  •  1  where  the  coil  pitch  is  180  —  a  electrical  degrees. 
4 

249.  Rating  of  Alternators. — Alternators  are  designed  to  give 
a  certain  terminal  voltage  and  to  supply  any  current  up  to  a 
certain  maximum  or  full-load  current. 

The  output  is 

P  =  nEI  cos  6  watts, 
where 

E  is  the  voltage  per  phase, 

/  is  the  full-load  current  per  phase, 

cos  6  is  the  power  factor  of  the  load,  and 

n  is  the  number  of  phases. 

The  power  output,  therefore,  depends  on  the  voltage  which  is  a 
fixed  quantity,  the  current  which  is  variable  and  is  limited  by  the 
allowable  temperature  rise  caused  by  the  copper  losses  and  other 
losses  in  the  machine,  and  the  power  factor  of  the  load  over  which 
the  designer  has  no  control. 

.  Alternators  should,  therefore,  be  rated  not  in  watts  or  kilowatts 
which  depend  on  the  power  factor  but  in  volt-amperes  or  kilovolt- 
amperes. 


SYNCHRONOUS  MACHINERY  301 

A  machine  rated  at  1,000  kva.  can  supply  1,000  kw.  to  a  non- 
inductive  load  at  unity  power  factor  or  it  can  supply  1,000  X 
0.80  =  800  kw.  to  an  inductive  load  of  80  per  cent,  power  factor. 

250.  Comparative  Ratings  of  an  Alternator  Wound  Single-, 
Two-  and  Three-Phase. — Take  the  case  of  a  machine  with  six 
slots  per  pole.  Let  e  be  the  effective  value  of  the  e.m.f.  generated 
in  each  coil  of  the  winding  and  I  be  the  current  per  conductor. 
The  current  will  be  the  same  in  the  three  cases  for  the  same  tem- 
perature rise. 

When  wound  single-phase  using  all  the  slots  the  distribution 
factor  is  0.64  and  the  terminal  e.m.f.  is 

E  =  6e  X  0.64 
and  the  output  is 

P'i  =  El  cos  </>  =  3.84e7  cos  <£, 

where  cos  $  is  the  power  factor  of  the  load. 

When  wound  single-phase  'using  only  four  slots  per  pole  the 
terminal  e.m.f.  is 

E  =  4e  X  0.83 
and  the  output  is 

Pi  =  El  cos  </>  =  3.3207  cos  <£. 

When  wound  two-phase  with  three  slots  per  phase  per  pole  the 
distribution  factor  is  0.91,  the  e.m.f.  per  phase  is 

E  =  3e  X  0.91 
and  the  output  is 

P2  =  2J5J7  cos  <f>  =  5.46e/  cos  <f>. 

When  wound  three-phase  with  two  slots  per  phase  per  pole  the 
distribution  factor  is  0.96,  the  e.m.f.  per  phase  is 

E  =  2e  X  0.96 
and  the  output  is 

P3  =  3EI  cos  0  =  5.76e/  cos  0. 

Taking  the  three-phase  rating  as  100  the  comparative  ratings 
are  as  given  below. 

Number  of  phases  Rating 

Three-phase  100.0 

Two-phase  95 . 0 

Single-phase  using  all  the  slots  67 . 0 

Single-phase  using  only  four  slots  per  pole  57 . 7 


302 


ELECTRICAL  ENGINEERING 


In  practice  an  alternator  is  given  the  same  rating  two-  and  three- 
phase  and  65  per  cent,  of  that  rating  single-phase. 

251.  Armature  Reaction. — The  flux  distribution  in  the  air  gap 
of  an  alternator  at  no  load  is  symmetrical  about  the  center  line  of 
the  pole  and  usually  follows  approximately  a  sine  wave.  The 
e.m.f.  generated  in  the  armature  is  also  a  sine  wave  (see  Fig. 
285). 

When  current  flows  in  the  armature  winding,  the  m.m.f.  of  the 
armature  combines  with  the  m.m.f.  of  the  field  and  changes  both 


CdH-^s 


77  <,T7 


Alte-uator 


Synchronous 
Motor 


S  I     Alternator 

FIG.  290. — Armature  reaction  and  reactance. 


the  magnitude  and  distribution  of  the  flux  crossing  the  air  gap 
and  cut  by  the  armature  conductors.  It  thus  changes  both  the 
magnitude  and  the  wave  form  of  the  e.m.f.  generated.  These 
results  are  termed  the  armature  reaction. 

Armature  reaction  depends  not  only  on  the  intensity  of  the 
current  in  the  armature  but  also  on  its  phase  relation  with  the 
generated  e.m.f.  Fig.  290  illustrates  armature  reaction  in  a 
machine  with  a  single-phase  concentrated  winding. 

In  (a)  the  armature  coil  is  shown  in  the  position  of  zero  e.m.f; 
if  the  current  is  in  phase  it  is  also  zero. 


SYNCHRONOUS  MACHINERY  303 

In  (b)  the  e.m.f.  is  maximum  and  the  current  is  maximum. 
The  m.m.f.  of  the  armature  is  cross-magnetizing,  that  is,  it  de- 
creases the  flux  over  one-half  of  the  pole  and  increases  it  over  the 
other  half.  The  useful  flux  is  only  decreased  by  the  small  amount 
lost  due  to  the  higher  saturation,  and,  therefore,  decreased  permea- 
bility over  the  half  of  the  pole  where  the  density  is  increased. 
The  flux  distribution  no  longer  follows  a  sine  wave  and  the  e.m.f. 
will  not  be  a  sine  wave. 

In  (c)  the  current  is  maximum  but  lags  90  degrees  behind  the 
generated  e.m.f.  The  m.m.f.  of  the  armature  acts  directly 
against  the  m.m.f.  of  the  field.  It  is,  therefore,  demagnetizing 
and  decreases  the  flux  but  does  not  distort  it. 

In  (d)  the  current  is  maximum  and  leads  the  e.m.f.  by  90 
degrees.  The  m.m.f.  of  the  armature  acts  directly  with  the  field 
and  magnetizes  it.  The  useful  flux  is  increased  and  is  not 
distorted. 

The  following  results  have  been  obtained : 

1.  A  current  in  phase  with  the  generated  e.m.f.   is  cross- 
magnetizing  and  only  decreases  the  flux  to  a  very  slight  extent. 

2.  A  current  lagging  90  degrees  behind  the  generated  e.m.f. 
demagnetizes  the  field  and  decreases  the  flux  and  decreases  the 
generated  e.m.f. 

3.  A   current   leading   the   generated   e.m.f.   by   90   degrees 
magnetizes  the  field,  increases  the  flux  and  increases  the  generated 
e.m.f. 

If  the  current  lags  behind  the  e.m.f.  by  angle  <f>,  it  may  be 
resolved  into  two  components,  /  cos  <j>  in  phase  with  the  e.m.f. 
and,  therefore,  cross-magnetizing  and  I  sin  </>  lagging  90  degrees 
behind  the  e.m.f.  and  demagnetizing. 

The  effect  of  armature  reaction  increases  almost  directly  with 
the  current  until  the  magnetic  circuit  becomes  saturated  after 
which  it  increases  much  faster  than  the  current.  Due  to  satura- 
tion the  cross-magnetizing  effect  increases  faster  than  the  current ; 
the  demagnetizing  m.m.f.  increases  directly  with  the  current  and 
the  decrease  of  flux  caused  by  it  would  be  proportional  to  the 
current,  if  it  were  not  for  the  change  in  the  leakage  factor  of  the 
machine.  The  increase  of  the  leakage  factor  under  load  is 
largely  due  to  the  presence  of  the  demagnetizing  armature  m.m.f. 
and  the  resulting  decrease  of  flux  must  be  charged  against  it. 

In  salient  pole  machines  the  cross-magnetizing  m.m.f.  acts  on 
a  path  of  much  larger  reluctance  than  the  demagnetizing  m.m.f. 


304  ELECTRICAL  ENGINEERING 

and  its  effect  is  correspondingly  smaller.  In  machines  with  cylin- 
drical rotors  the  reluctance  of  the  air  gap  is  uniform  over  the 
whole  periphery  and  the  two  m.m.fs.  therefore  act  on  similar 
paths. 

252.  Armature  Reactance. — The  flux  produced  by  the  cur- 
rent in  the  armature  coil  in  Fig.  290  may  be  separated  into  two 
parts  as  shown. 

Part  (1)  is  the  flux  of  armature  reaction  which  crosses  the  gap 
and  interferes  with  the  flux  threading  the  field  circuit.  Its  effect 
is  either  cross-magnetizing,  demagnetizing  or  magnetizirig. 

Part  (2)  is  the  flux  which  only  interlinks  with  the  coil  itself  and 
does  not  interfere  with  the  flux  produced  by  the  field  m.m.f.  It 
is  the  self-inductive  flux  of  the  coil  and  generates  in  the  coil  an 
e.m.f.  of  self-inductance,  which  consumes  a  component  of  the 
e.m.f.  generated  by  rotation.  This  e.m.f.  called  the  armature 
reactance  drop  is  equal  to  the  product  of  the  armature  current  I 
and  the  armature  reactance  x  and  leads  the  current  by  90  degrees. 

The  reactance  is  x  =  2irfL,  where  L  is  the  inductance  of  the 
armature.  L  and  x  both  decrease  as  the  armature  current  in- 
creases due  to  the  increased  saturation  and,  therefore,  decreased 
permeability  of  the  leakage  path  surrounding  the  armature  con- 
ductors. They  also  vary  as  the  armature  is  rotated,  when  the 
conductor  is  under  the  pole  the  reluctance  of  its  local  leakage 

/Tooth   Tip   Leakage  Flux 
Tooth  Tip  Leakage  Flux 


One  Slot   per   Phase  per  Pole  Three   Slots  per  Phase  per  Pole 

FIG.  291. — Slot  leakage  flux  and  tooth- tip  leakage  flux. 

path  is  minimum  and  L  and  x  are  large,  when  between  the  poles 
the  reluctance  is  maximum  and  L  and  x  are  reduced.  An  aver- 
age value  of  x  is  chosen  to  represent  the  armature  reactance.  At 
light  loads  when  the  current  is  small  the  reactance  drop  will  be 
greater  than  the  value  corresponding  to  the  average  reactance 
and  when  the  current  is  large  it  will  be  smaller. 

253.  Leakage  Fluxes. — The  leakage  fluxes  of  armature  react- 
ance may  be  separated  into  three  parts:  (a)  the  slot  leakage  flux, 
(b)  the  tooth-tip  leakage  flux,  Fig.  291,  and  (c)  the  end  connection 
leakage  flux,  Fig.  292. 


SYNCHRONOUS  MACHINERY 


305 


(a)  The  slot  leakage  flux  is  directly  proportional  to  the  number 
of  ampere  conductors  in  the  slot,  that  is,  to  the  total  current  in  the 
slot;  it  increases  directly  with  the  slot  depth  and  is  inversely 
proportional  to  the  slot  width.  If  the  slots  are  partly  closed,  the 
component  of  the  flux  across  the  narrow  opening  is  very  large. 
The  slot  leakage  flux  is  independent  of  the  position  of  the  slot 
relative  to  the  pole;  due  to  saturation  of  the  teeth  it  does  not 
increase  directly  with  the  current  and  therefore  the  component 
of  the  armature  reactance  due  to  it  decreases  with  increase  of 
current. 

(6)  The  tooth  tip  leakage 
follows  paths  such  as  those 
shown  in  Fig.  291.  It  cannot 
be  calculated  so  easily  or  so 
accurately  as  the  slot  leakage 
and  it  depends  to  a  limited 
extent  on  the  position  of  the 


FIG.  292. — End-connection  leakage  flux. 


slot  relative  to  the  pole,  espe- 
cially if  the  air  gap  is  short. 
This  does  not  apply  to  ma- 
chines with  cylindrical  rotors. 

(c)  The  end-connection 
leakage  flux  follows  a  path  entirely  in  air.  It  is  proportional  to 
the  number  of  ampere  conductors  in  the  phase  belt  but  when 
the  winding  is  distributed  in  a  number  of  slots  the  length  of  the 
leakage  path  is  increased.  The  great  length  of  the  end  connec- 
tions makes  this  component  of  the  leakage  flux  comparatively 
large.  Since  the  path  never  becomes  saturated  this  flux  is 
directly  proportional  to  the  current.  Distributing  the  winding 
in  a  number  of  slots  per  phase  per  pole  decreases  all  of  these 
leakage  fluxes  by  increasing  the  length  of  the  leakage  paths. 

254.  Polyphase  Armature  Reaction. — If  n  is  the  number  of 
turns  per  phase  per  pair  of  poles  on  a  two-phase  alternator  and 
i\  =  Im  cos  6  is  the  current  in  phase  1  and  iz  =  Im  cos  (0  —  90°) 
=  Im  sin  6  is  the  current  in  phase  2,  the  m.m.fs.  of  the  two  phases 
are  mi  =  ra'i  =  nlm  cos  6  and  m2  =  nlm  sin  0.  The  two  m.m.fs. 
are  in  quadrature  in  time  and  space  but  combine  to  give  a  con- 
stant m.m.f.  nlm  fixed  in  position  relative  to  the  field  m.m.f. 
and  revolving  synchronously  backward  relative  to  the  armature. 
This  can  be  seen  by  reference  to  Fig.  293.  AB  is  the  winding 
of  phase  1  and  the  current  is  assumed  to  lag  behind  the  e.m.f.  by 
20 


306 


ELECTRICAL  ENGINEERING 


angle  <£.  At  the  instant  represented  the  current  is  maximum  and 
the  m.m.f.  of  the  coil  is  maximum  and  acts  in  direction  OF.  The 
current  in  phase  2  is  now  zero.  The  m.m.fs.  acting  are  shown  in 
Fig.  294. 


FIG.  294. 


In  Fig.  295  the  coil  A  B  has  moved  through  angle  6  and  its  cur- 
rent has  decreased  to  Im  cos  6  and  its  m.m.f.  to  nlm  cos  6.  The 
current  in  coil  CD  has  a  value  Im  sin  8  and  its  m.m.f.  is  nlm  sin  6. 


FIG.  295. 


FIG.  296. 


The  component  of  the  m.m.f.  of  phase  1  in  direction  OY  is 
nlm  cos  B  -  cos  6  =  nlm  cos2  6  and  the  component  in  direction  O^Tis 
—  nlm  cos  6-  sin  0. 

The  component  of  the  m.m.f.  of  phase  2  in  direction  OF  is 
nlm  sin  B  -  sip-  3  =  nlm  sin2  6  and  the  component  in  direction  OX  is 


SYNCHRONOUS  MACHINERY 


307 


-f-  nlm  sin  0-cos  6.  The  resultant  m.m.f.  of  the  two  phases  in  di- 
rection OF  is  nlm  cos2  0  +  nlm  sin2  0  =  nlm  and  in  the  direction 
OX  is  nlm  cos  6  -  sin  0  —  nlm  cos  0  •  sin  0  =  0. 

Thus  the  resultant  armature  m.m.f.  is  nlm  in  fixed  direction 
relative  to  the  field  m.m.f.  and,  therefore,  revolving  synchron- 
ously relative  to  the  armature  (Fig.  296). 

The  direction  of  the  resultant  armature  m.m.f.  relative  to  the 
field  m.m.f.  is  determined  by  the  angle  of  phase  difference  between 
the  current  and  the  e.m.f.  generated  at  no  load. 

If  the  current  is  in  phase  with  the  e.m.f.,  the  armature  m.m.f. 
acts  at  right  angles  to  the  field  m.m.f.  and  is,  therefore,  cross- 
magnetizing  only;  if  the  current  lags  by  angle  $,  the  armature 
m.m.f.  can  be  separated  into  two  components  nlm  sin  </>  which  is 
demagnetizing  and  nlm  cos  <f>  which  is  cross-magnetizing  (Fig. 
294). 

Y 


Cos(0-240) 
X 


FIG.  297. 


FIG.  298. 


If  the  two-phase  winding,  Fig.  295,  is  replaced  by  a  three-phase 
winding,  Fig.  297,  with  the  first  phase  AB  in  the  same  position 
as  before  and  the  other  phases  CD  and  EF  displaced  120  degrees 
and  240  degrees  from  it,  the  m.m.fs.  of  the  three  phases  will  be 
respectively  nlm  cos  0,  nlm  cos  (0  —  120)  and  nln  cos  (0  —  240) 
and  will  act  in  the  directions  represented.  As  before  0  is  meas- 
ured from  the  instant  of  maximum  current  and  the  currents  in 
the  three  phases  are  assumed  to  lag  behind  the  e.m.fs.  by  angle  </>. 

The  sum  of  the  components  of  m.m.f.  in  direction  OF  is 

nlm  cos20  +  nlm  cos2  (0  -  120)  +  nlm  cos2  (0  -  240)  =  %nlm. 
and  the  sum  of  the  components  in  the  direction  OX  is 

nlm  cos  0  sin  0    -f  nlm  cos  (0  -  120)  sin  (0  -  120) 

+  nlm  cos  (0  -  240)  sin  (0  -  240)  =  0. 


308 


ELECTRICAL  ENGINEERING 


Thus,  the  resultant  m.m.f.  of  the  armature  of  a  three-phase  alter- 
nator is 

Ma=%nlm,  (289) 

where  n  is  the  number  of  turns  in  series  per  phase  and  Im  is  the 
maximum  value  of  the  armature  current. 

The  armature  m.m.f.  is  fixed  in  direction  relative  to  the  fields 
and  revolves  synchronously  relative  to  the  armature  (Fig.  298). 


©©©©©@®®®®®©©@(B©© 

w  ^rn^r 


flTTp 


44 


ll-0.86lm    C 

®r.°..8,I«TTTj 

Time, «  =  £._[_     _Q 


HUH 


S.O 


©©® 


®Q 


Armature  M.M.F.with  the  Current  in  Phase  with    js 
the  Generated  Voltages  per  Phase.    Cross-MagnetlzingOH 

*i  =  °    .  Q©QO®®®®OO©©©©OO®® 

2.2 —~  v«OOlW         *       M        *  *       ^     "^  *       Kl        ' 

Time,  ^*0 


HUH 


HI  IH       5 


Armature  M.M.F.with  the  Current  in  Quadrature 
Behind  the  Generated  Voltages  per  Phase.  Demagnetizing] 


Voltages  per  Phase 


urrents  per  Phase 
(in  Phase) 


JOC 

Currents  per  Phase 
(Lagging) 


HUH 


a        Currents  per  Phase 
0  (Leading) 


Armature  M.M.F.with  the  Current  in  Quadrature 
Ahead  of  the  Generated  Voltages  per  Phase.    Magnetizing 

FIG.  299. — Armature  m.m.f.  of  a  three-phase  alternator. 


Fig.  299  shows  the  same  facts  in  another  way  in  the  case  of  a 
three-phase  alternator  with  its  winding  distributed  in  six  slots 
per  pole. 

(1)  is  a  section  through  the  conductors  indicating  the  directions 
and  intensities  of  the  e.m.fs.  for  the  time  t  =  0. 


SYNCHRONOUS  MACHINERY 


309 


(2)  shows  the  e.m.f.  waves  generated  in  the  three  phases. 

(3)  shows  the  armature  winding. 

(4),  (5)  and  (6)  represent  the  armature  m.m.fs.  corresponding 
to  the  three  instants,  t  =  0,  t  =  ti  and  t  =  t2  or  6  =  0,  0  =  30 
degrees  and  6  =  60  degrees,  with  the  currents  (7)  in  phase  with 
the  generated  e.m.f.  The  armature  m.m.f.  is  cross-magnetizing; 
its  value  represented  by  the  area  under  the  m.m.f.  lines  is  approx- 
imately constant  and  it  is  fixed  in  position  relative  to  the  poles 
and  therefore  revolves  relative  to  the  armature. 

(8)  and  (9)  show  the  armature  m.m.f.  and  currents  at  time  t  = 
0,  with  the  currents  in  quadrature  behind  the  e.m.fs.  and  there- 
fore demagnetizing. 

(10)  and  (11)  show  the  armature  currents  leading  and  the 
armature  m.m.f.  magnetizing, 

255.  Single -phase  Armature  Reaction. — In  a  single-phase 
alternator  with  n  armature  turns  per  pair  of  poles  carrying  a 
current  i  =  Im  sin  0,  the  armature  m.m.f.  varies  from  a  maximum 
value  nlm  to  zero;  it  is  fixed  in  direction  relative  to  the  armature 
and  revolves  relative  to  the  poles.  It  produces  a  double-fre- 
quency pulsation  of  the  field  and  a  third  harmonic  of  e.m.f. 


® 


FIG.  300. — Single-phase   armature  reaction. 


Fig.  300  shows  four  instants  during  the  revolution  of  a  single- 
phase  alternator  in  which  the  current  is  assumed  to  be  in  quadra- 
ture behind  the  generated  e.m.f.  In  (1)  the  e.m.f.  is  zero  and 
the  current  is  maximum,  the  armature  m.m.f.  is  demagnetizing 
and  reduces  the  flux.  In  (2)  the  current  is  zero  and  the  flux  has 
its  maximum  value.  In  (3)  the  current  is  maximum  again  and 
the  flux  is  reduced  as  in  (1).  In  (4)  the  current  is  zero  again. 

The  values  of  the  flux  crossing  the  gap  for  one  revolution  or 
one  cycle  are  shown  in  Fig.  301.  The  flux  may  be  separated  into 
two  parts,  the  constant  value  $  and  the  alternating  flux  of 
amplitude  $1,  which  goes  through  two  complete  cycles  during  one 
cycle  of  the  current  or  e.m.f.  Single-phase  armature  reaction 


310 


ELECTRICAL  ENGINEERING 


thus  produces  a  double-frequency  pulsation  of  the  flux  crossing 
the  gap. 

If  time  t  and  angular  displacement  0  of  the  rotating  field  are 
measured  from  position  (1),  the  flux  inclosed  by  the  coil  at  this 
instant  may  be  represented  by 

<£o  =  $  —  0i  cos  26. 


1 

/^\           /0X 

/     V 

<  1  Flux,, 

7A 

4 

rent 

>  x 

& 

0, 

T 

0min 

Cycle 
1  Cur 

Cycle 
D   ©     ®    @ 

Resultant 
E.M.F.  Wave 


SO       90      180     270     360 

h     Angular  Displacement  =  0 

FIG.  301.  FIG.  302. 

At  time  t  and  .angle  0  after  position  (1)  the  flux  inclosed  is 
<t>  =  4>o  cos  6  =  $  cos  0—<£i  cos  20  •  cos  0 

=  $  cos  0  -  ^  (cos  30  +  cos  0) 

and  the  e.m.f.  generated  in  the  coil  is 


d<f>  d 

6"      ~Udt    =  ~ndt 


cos  0  -  y  (cos  30  +  cos  0)  1 


=  n  \  3>  sin  0  —  ~  (3  sin  30  +  sin  0)  I  -^ 
2  }  at 

but  0  =  2wft  and  —  =  2irf. 
and  therefore 

e  =  2-n-fn  j  ( <J>  —  ~j  sin  0  —  %  <f>i  sin  30 
I  \  2  / 

i  in 

=  E    sin  d  -  E    sin  30. 

m  m 

The  generated   e.m.f.    consists   of  an   e.m.f.  of  fundamental 
frequency  having  a  maximum  value  Em  =  2irfn  ( <£  —  ~)  and  a 

third  harmonic  of  maximum  value  E      =  Sirfn  0ij  the  twoe.m.fs. 

m 

pass  through  zero  together  but  in  opposite  directions  and  the 
resultant  wave  is  symmetrical  with  a  peak  at  the  center  (Fig.  302) . 
Thus  single-phase  armature  reaction  produces  a  double-fre- 
quency pulsation  of  the  field  and  a  third  harmonic  of  e.m.f. 


SYNCHRONOUS  MACHINERY        311 

Since  the  magnetic  circuit  is  surrounded  by  a  field  winding, 
consisting  of  a  large  number  of  turns,  the  pulsations  will  be  less 
than  the  above  results  indicate.  The  variation  of  the  flux  linking 
with  the  field  winding  induces  in  it  e.m.fs.  and  currents  which 
oppose  the  variation  and  limit  it  to  a  small  value. 

In  a  machine  with  a  large  number  of  field  turns  the  pulsation 
produced  by  armature  reaction  up  to  full-load  current  is  very 
small  and  the  armature  reaction  may  be  considered  as  constant  in 
value  with  reference  to  the  fields. 

In  the  case  of  a  short-circuit,  however,  where  from  three  to  five 
times  full-load  current  flows  in  the  armature,  a  large  pulsation  of 
flux  is  produced  in  the  magnetic  circuit  and  very  large  e.m.fs. 
and  currents  may  be  induced  in  the  field  windings. 

If  a  short-circuit  occurs  on  one  phase  only  of  a  three-phase 
alternator,  the  armature  reaction  can  be  separated  into  the  ordi- 
nary three-phase  armature  reaction  with  equal  currents  and  a 
single-phase  armature  reaction  due  to  the  excess  of  the  short- 
circuit  current  over  normal  current  acting  in  the  turns  of  one 
phase.  This  single-phase  armature  reaction  produces  a  double- 
frequency  pulsation  of  the  field  and  a  third  harmonic  of  e.m.f. 
in  all  the  phases. 

The  effects  of  single-phase  armature  reaction  are  relatively 
greater  in  machines  with  a  small  number  of  turns  on  the  fields,  as 
turbo  alternators. 

256.  Electromotive  Forces  in  the  Alternator. — In  studying 
the  performance  of  an  alternator  it  is  necessary  to  determine  the 
relation  between  the  terminal  e.m.f.  E,  the  e.m.f.  E\  generated  by 
rotation  and  e.m.f.  EQ  generated  at  no  load. 

EI  is  the  e.m.f.  generated  in  the  armature  by  the  rotation  of  the 
flux  produced  in  the  air  gap  by  the  resultant  of  the  m.m.fs.  of 
the  field  and  armature.  It  is  the  vector  sum  of  the  terminal 
e.m.f.  E  and  the  e.m.f.  consumed  by  the  impedance  of  the  arma- 
ture. The  armature  impedance  is  Z  =  \/r2  +  x2,  or  expressed 
in  rectangular  coordinates  Z  =  r  +  jx,  where  r  is  the  resistance 
of  the  armature  and  consumes  a  component  of  e.m.f.  Ir  in 
phase  with  the  current  /,  and  x  is  the  true  self-inductive  reactance 
of  the  armature  and  consumes  a  component  of  e.m.f.  Ix  in  quad- 
rature ahead  of  the  current. 

The  generated  e.m.f.  thus  is 

&  =  £  +  IZ 

=  $  +  I(r+jx),  (290) 


312  ELECTRICAL  ENGINEERING 

and  the  terminal  e.m.f .  is  the  vector  difference  between  the  e.m.f. 
generated  in  the  armature  by  rotation  and  the  impedance  drop 

E  =  Et-  I  (r+jx).  (291) 

Eo  is  the  e.m.f.  generated  at  no  load  due  to  cutting  the  flux  pro- 
duced by  the  field  m.m.f.  Mf  acting  alone.  Under  load  current 
flows  in  the  armature  and  exerts  a  m.m.f.  Ma,  which  is  either 
cross-magnetizing,  demagnetizing  or  magnetizing  depending  on 
the  phase  relation  of  the  current  and  the  terminal  e.m.f.  This 
armature  m.m.f.  combines  with  the  field  m.m.f.  and  changes  both 
the  intensity  and  the  distribution  of  the  flux  in  the  gap,  so  that 
under  load  the  e.m.f.  generated  in  the  armature  is  not  the  same  as 
at  no  load.  •  . 

The  difference  between  the  two  is  the  e.m.f.  consumed  or  the 
e.m.f.  not  generated  due  to  the  presence  of  the  armature  reaction. 
This  e.m.f.  is  proportional  to  the  current  and  can  be  expressed  as 
the  product  of  the  current  /  and  a  component  of  reactance  x' '.  It 
is  Ix'  and  is  in  quadrature  ahead  of  the  current.  Thus 

EQ  =  El  +  jlx' 

=  $  +  I(r+jx)  +  jlxf 

=  E  +  I{r+j(x  +  x')} 

=  E  +  I(r+jx8).  (292) 

The  total  reactance  of  the  armature  xs  is  called  the  synchronous 
reactance  and  consists  of  two  components,  x  which  represents  the 
effect  of  the  armature  leakage  flux  and  which  has  been  called  the 
armature  reactance  and  x'  which  represents  the  effect  of  armature 
reaction. 

The  quantity  Z8  =  r  +  jx8  is  the  synchronous  impedance  of  the 
armature  and  consumes  a  component  of  the  no-load  e.m.f.  IZ8  = 
I  (r  +  fas)  which  is  the  synchronous  impedance  drop  in  the 
armature. 

Thus  the  e.m.f.  generated  at  no  load  is  the  vector  sum  of  the 
terminal  e.m.f.  and  the  synchronous  impedance  drop 

EQ  =  E  +  I(r+jxs). 

257.  Armature  Resistance. — The  resistance  of  the  armature  r, 
used  in  calculations  for  regulation  and  efficiency,  is  not  the  true 
ohmic  resistance  as  measured  by  passing  direct  current  through 
the  winding  but  is  from  20  to  100  per  cent,  greater  than  this 
value.  This  increased  resistance  is  sometimes  called  the  effective 
resistance  of  the  armature. 


SYNCHRONOUS  MACHINERY 


313 


The  loss  of  power  which  is  charged  against  the  armature  copper 
is  the  increase  in  the  total  losses  in  the  machine  due  to  the  presence 
of  the  armature  winding  carrying  current.  This  loss  includes  (1) 
the  true  72r  loss  due  to  the  passage  of  the  load  current,  (2)  any 
increase  in  the  iron  losses  due  to  the  distortion  of  the  flux  by  the 
armature  currents  and  (3)  eddy-current  losses  in  the  armature 
copper,  such  as  those  discussed  in  Art.  188. 

In  ordinary  cases  it  is  advisable  to  add  about  50  per  cent,  to 
the  resistance  measured  by  direct  current;  a  slight  error  will  not 
affect  the  efficiency  of  the  machine  to  any  great  extent  since  the 
copper  loss  is  usually  considerably  less  than  the  iron  losses. 

258.  Vector  Diagram  of  Electromotive  Forces  and  Magneto- 
motive Forces. — In  Fig.  303  (a)  is  shown  the  vector  diagram  of  an 
alternator  supplying  an  inductive  load. 


FIG.  303. 


E  =  terminal  e.m.f. 

7  =  armature  current  lagging  behind  E  by  angle  <j>. 

IT  =  e.m.f.  consumed  by  armature  resistance,  in  phase 
with  7. 

Ix  =  e.m.f.  consumed  by  armature  reactance,  in  quadra- 
ture ahead  of  the  current. 

Eb  =  —  Ix  =  counter  e.m.f.  of  armature  reactance  due 
to  the  armature  leakage  flux  <£/,.  Eb  is  in  quadrature 
behind  fa  and  7. 

EI  =  e.m.f.   generated  by  rotation,  that  is,   the  e.m.f. 
generated  due  to  cutting  the  flux  produced  in  the 
magnetic  circuit  by  the  resultant  of  the  m.m.fs. 
of  field  and  armature,  EI  =  %  +  7  (r  +  jx)- 
M  a—  armature  m.m.f.,  in  phase  with  7. 


314 


ELECTRICAL  ENGINEERING 


M  =  resultant  of  the  field  m.m.f.  and  armature  m.m.f., 
in  quadrature  ahead  of  the  e.m.f.  EI  due  to  it. 

Mf  =  field  m.m.f.,  Mf  +  M a  =  M. 

E0  =  e.m.f.  generated  at  no  load  by  the  field  m.m.f.  act- 
ing  alone,  in  quadrature  behind  Mf. 

Ix'  =  EiEQ  =  component  of  the  no-load  e.m.f.  E0  con- 
sumed by  armature  reaction. 

Ix9  =  Ix  +  Ix'  =  I  (x  +  x'}  =  e.m.f.  consumed  by  ar- 
mature synchronous  reactance. 

IZ8  =  EE0  =  e.m.f.  consumed  by  synchronous  impedance, 

0o  =  angle  of  lag  of  the  current  7  behind  the  no-load 

generated  e.m.f.  E0. 

Ma  sin  0o  =  demagnetizing  component  of  the  armature  m.m.f. 
Ma  cos  0o  —  cross-magnetizing    component    of    the    armature 

m.m.f. 
This  diagram  may  be  simplified  to  that  shown  in  (6). 


Non-inductive  Load,  $=0 

FIG.  304. 


»         lr 

Inductive  Load,  0=60  Capacity  Load,  0=60 

Power  Factor  =  Power  Factor  = 

j     50  Percent  Lagging  50  Percent  Leading 


FIG.  305. 


FIG.  306. 


In  Figs.  304,  305  and  306  are  drawn  the  diagrams  for  the  three 
cases:  (1)  0  =  0,  or  non-inductive  load;  (2)  0  =  60  degrees  lag, 
or  inductive  load  with  a  power  factor  of  50  per  cent.;  and  (3) 
0  =  60  degrees  lead,  or  a  capacity  load  with  a  power  factor  of 
50  per  cent.  The  same  value  of  armature  current  is  taken  in  the 
three  cases. 

Referring  to  these  diagrams  it  may  be  seen  that  for  the  same 
terminal  voltage  and  the  same  armature  current  much  larger 
values  of  EQ  and  M/  are  required  for  inductive  loads  than  for 
non-inductive  loads  and  much  smaller  values  for  capacity  loads 
than  for  non-inductive  loads. 

While  the  magnetic  circuit  of  the  machine  is  unsaturated  the 
field  excitation  required  to  produce  the  e.m.f.  E0  is  approximately 


SYNCHRONOUS  MACHINERY  315 

proportional  to  it,  but  higher  up  on  the  saturation  curve  the 
required  excitation  increases  faster  than  the  e.m.f  .  The  relation 
between  the  field  current  or  field  m.m.f.  and  the  no-load  e.m.f. 
can  be  obtained  by  reference  to  the  no-load  saturation  curve  of 
the  alternator  in  Fig.  310. 

The  relation  between  E0,  E  and  7  can  be  expressed  algebraically 
as  indicated  in  Fig.  307  which  is  a  reproduction  of  Fig.  303(6). 

E0  =  od  =  Vo62  +  bd2  =  Voa  +  alz  +  be  +  'cd2 


=  V(E  cos  <£  +  7r)2  +  (E  sin  0  + 


a    b 
FIG.  307. 

The  same  result  may  be  obtained  by  expressing  the  e.m.fs. 
in  rectangular  coordinates  taking  the  current  /  as  real  axis. 
The  terminal  e.m.f.  is 

E  =  E  cos  <i>  +  jE  sin  <£; 
the  synchronous  impedance  drop  is 

/Z.  =  Ir+jlx8; 
and  the  no-load  e.m.f.  is 

E  =  E  +  !?8  =  (E  cos  0  +  Ir)  +  j  (E  sin  0  +  la?.); 
or  taking  the  absolute  value 


EQ  =      (E  cos  (^  +  7r)2  +  (E  sin  0  +  Ixs)2.          (293) 

259.  Voltage  Characteristics.—  The  relation  between  the  ter- 
minal e.m.f.  and  armature  current  of  an  alternator,  with  a  fixed 
value  of  field  current  and  a  given  load  power  factor,  is  called 
the  "  regulation  curve"  or  "voltage  characteristic"  for  the  given 
power  factor. 

When  the  power  factor  is  unity  and  the  current  is  in  phase  with 
the  terminal  e.m.f.  E}  it  lags  by  an  angle  <£0  (Fig.  304)  behind  the 


316 


ELECTRICAL  ENGINEERING 


no-load  e.m.f.  E0  and  there  is  thus  a  small  demagnetizing  effect 
proportional  to  /  sin  <£0,  which  decreases  the  flux  and  a  large  cross- 
magnetizing  effect  proportional  to  /  cos  fa  which  changes  the 
distribution  of  the  flux  but  only  decreases  it  slightly  due  to  satura- 
tion. Thus  even  with  non-inductive  load  the  armature  reaction 
causes  a  decrease  in  the  flux  crossing  the  air  gap,  and  the  e.m.f.  EI 
generated  by  rotation  is  less  than  the  no-load  e.m.f.  E0.  In 
addition,  the  armature  reactance  x  and  the  resistance  r  both  con- 
sume components  of  e.m.f.  proportional  to  the  current. 


120 


10.     20      30      40      50      60      70      80      90    100 
Percent  Load  Current 


0       10      20      30      40     50      60      70      80      90    100 
'Percent  Load  Current 


FIG.  308. — Voltage  characteristics  of       FIG.  309. — Compounding  curves  of  an 
an  alternator.  alternator. 


Therefore,  at  non-inductive  load  the  terminal  e.m.f.  falls  with 
increasing  current  as  shown  in  curve  (1),  Fig.  308,  which  is  the 
voltage  characteristic  for  unity  power  factor. 

With  inductive  load  the  demagnetizing  effect  is  increased  and 
the  terminal  e.m.f.  falls  off  more,  curve  (2).  With  a  capacity 
load  in  which  the  current  leads  the  terminal  e.m.f.  the  armature 
m.m.f.  is  magnetizing  and  so  raises  the  terminal  e.m.f.,  curve  (3). 

These  voltage  characteristics  are  calculated  from  equation 
(293)  on  page  315.  The  value  of  field  current  //  is  chosen  and  the 
corresponding  value  of  EQ  obtained  from  Fig.  310.  Any  required 
power  factor  cos  <j>  is  taken,  the  current  I  is  varied  and  the  values 
of  E  obtained  and  plotted  as  ordinates. 


SYNCHRONOUS  MACHINERY 


317 


260.  Compounding  Curves. — The  "compounding  curves"  or 
" field  characteristics"  show  the  relation  between  the  field  current 
and  armature  current  for  a  constant  terminal  e.m.f.  at  any  re- 
quired power  factor. 

Fig.  309  shows  the  compounding  curves  for  unity  power  factor, 
curve  (1),  80  per  cent,  power  factor  lagging,  curve  (2),  and  80  per 
cent,  power  factor  leading,  curve  (3). 

At  non-inductive  load  an  increase  of  field  current  is  required  as 
the  load  current  increases  to  maintain  a  constant  terminal  e.m.f. 

With  inductive  load  a  much  larger  increase  of  field  current  is 
required  to  counteract  the  effect  of  the  lagging  current. 

With  capacity  load  the  field  current  must  be  decreased  in  order 
to  maintain  a  constant  terminal  e.m.f. 

The  same  results  are  shown  in  the  diagrams,  Figs.  304,  305,  and 
306. 

These  compounding  curves  can  also  be  calculated  from  equation 

Eo  =  \/(E  cos  0  +  Ir)2  +  (E  sin  0  +  Ixs).z 


a    b         c 

Field  Kxcitatiou 


field  Ejcitation   \f  ™  My 

FIG.  310. — Saturation  curves. 

E  remains  constant,  a  certain  value  of  cos  <f>  is  chosen,  /  is 
varied  and  the  value  of  Eo  is  calculated.  The  corresponding  value 
of  //  is  obtained  from  the  saturation  curve,  Fig.  310,  and  is 
plotted  on  the"  armature-current  base. 

261.  Tests  for  the  Determination  of  the  Regulation  of  Alter- 
nating-current Generators. — The  standards  committee  of  the 
American  Institute  of  Electrical  Engineers  suggests  the  three 
following  methods  of  determining  the  regulation  of  alternating- 
current  generators.  They  are  given  in  the  order  of  preference. 


318 


ELECTRICAL  ENGINEERING 


Method  (a). — The  regulation  may  be  measured  directly  by 
loading  the  generator  with  the  specified  load  and  power  factor 
and  then  reducing  the  load  to  zero.  The  difference  between  the 
voltage  readings  at  no  load  and  full  load  expressed  as  a  per  cent, 
of  the  full-load  voltage  is  the  regulation  under  the  specified 
conditions.  The  two  voltage  readings  must  be  taken  under  the 
same  conditions  of  speed  and  excitation. 

This  method  cannot  be  applied  generally  for  shop  tests  on 
large  generators  and  either  method  (6)  or  (c)  must  be  used. 

Method  (6). — The  regulation  is  determined  from  data  obtained 
from  the  no-load  saturation  curve  and  the  full-load  saturation 
curve  for  zero  power  factor  (Fig.  310).  The  latter  curve  may  be 
obtained  by  loading  the  generator  with  under-excited  synchron- 
ous motors,  which  can  be  made  to  give  a  very  low  power  factor. 

From  these  two  curves  points  on  the  load  saturation  curves 
for  any  other  power  factor  can  be  obtained  as  follows:  With 


FIG.  311. 

excitation  oa  the  open-circuit  voltage  is  am  =  EQ  and  the  termi- 
nal voltage  with  full-load  current  at  zero  power  factor  is  aA:3. 
The  drop  of  voltage  is  wfc3  and  this  is  the  voltage  consumed  by 
the  synchronous  reactance  of  the  armature,  since  under  the 
condition  of  zero  power  factor  it  is  directly  subtracted  from  the 
open-circuit  voltage  if  the  resistance  drop  is  neglected. 
Fig.  311  shows  the  vector  diagram  for  this  case. 

EQ  =  am  =  no-load  or  open-circuit  voltage. 

/  =  full-load  current  in  quadrature  behind  EQ. 
Ix8  =  voltage   consumed   by   the   synchronous   reactance   in 

quadrature  ahead  of  /. 

E  =  akz  =  terminal  voltage;  it  is  obtained  by  subtracting 
=  Ix8  from  EQ.    * 


SYNCHRONOUS  MACHINERY 


319 


To  obtain  the  corresponding  point  on  the  load  saturation  curve 
for  any  other  power  factor  the  vector  k3m  —  Ixa  must  be  sub- 
tracted from  EQ  in  its  proper  phase  relation. 

The  diagram  in  Fig.  3 12  (a)  may  be  applied  for  any  power 
factor.  The  horizontal  line  through  a  is  taken  as  the  direction 
of  the  constant  current  vector  /  and  a  circle  of  radius  am  =  EQ 
is  described  about  a.  The  terminal  voltage  line  for  any  power 
factor  cos  4>  makes  an  angle  <£  with  the  current  line;  the  voltage 
consumed  by  synchronous  reactance  as  =  Ixs  is  se.t  up  in  quadra- 
ture ahead  of  I  and  through  its  extremity  s  a  line  is  drawn  parallel 
to  the  terminal  voltage  to  cut  the  circle  of  radius  EQ  aim.  From 
m  the  line  mks  is  drawn  parallel  to  as  to  cut  the  terminal  voltage 
line  at  &3.  The  vector  ak$  then  represents  the  terminal  voltage 
E  under  the  given  conditions. 

This  diagram  may  be  replaced  by  the  simplified  diagram  in 
Fig.  312(6)  which  gives  the  same  results.  The  circle  of  radius 
Eo  is  drawn  about  m  as  center  instead  of  a. 

If  it  is  desired  to  take  account  of  the  resistance  drop  the  dia- 
grams in  Fig.  312  may  be  replaced  by  those  in  Fig.  313. 


(a) 


FIG.  313. 


In  Fig.  310  load  saturation  curves  for  unity  power  factor 
(0  =  0)  and  80  per  cent,  power  factor  (<f>  =  36.8  degrees)  are 
shown. 

The  values  of  regulation  corresponding  to  the  various  power 
factors  and  with  a  fixed  field  excitation  oa  are  as  follows  : 


Regulation  at  unity  power  factor          =  ~ 
Regulation  at  80  per  cent,  power  factor  =  — 
Regulation  at  zero  power  factor  — 


100  per  cent.  =       lOO  per  cent.  =  8.7   per  cent. 


100  per  cent.  =  |^100  per  cent.  =  26.5  per  cent. 
*—  100  per  cent.  =  ^100  per  cent.  =  43.0  per  cent. 


320 


ELECTRICAL  ENGINEERING 


When  expressing  the  regulation  of  a  generator  for  various 
power  factors  it  is  usual  to  make  the  comparison  on  a  basis  of 
equal  terminal  voltages  =  rated  voltage  and  not  under  the  condi- 
tion of  fixed  excitation. 

To  obtain  these  values  of  regulation  the  various  saturation 
curves  must  be  extended  to  cut  the  rated  voltage  line. 


The  regulation  at  unity  power  factor 
The  regulation  at  "80  per  cent,  power  factor 
The  regulation  at  zero  power  factor 


=  —  -£  100  per  cent.  =  7  per  cent. 


100  per  cent.  =16  per  cent. 
100  per  cent.  =  25  per  cent. 


These  values  could  not  be  obtained  from  a  single  vector  dia- 
gram since  the  synchronous  reactance  drop  is  not  constant  but 
decreases  as  the  saturation  increases. 


FlG.    314. 

Method  (c). — If  it  is  not  possible  to  obtain  the  full-load  satura- 
tion curve  at  zero  power  factor  from  test,  it  may  be  constructed 
from  the  no-load  saturation  curve  and  the  short-circuit  curve 
(Fig.  314).  The  short-circuit  curve  is  obtained  by  short-circuit- 
ing the  generator  through  an  ammeter.  A  very  low  value  of 
field  excitation  must  be  taken  at  start  and  gradually  increased 
until  the  armature  current  is  about  twice  full-load  current.  The 


SYNCHRONOUS  MACHINERY        321 

curve  showing  the  relation  between  the  armature  current  and 
the  field  current  is  a  straight  line  passing  through  the  origin  and 
it  may  be  extended  to  any  desired  length. 

Since  the  armature  is  short-circuited  and  there  is  therefore  no 
terminal  voltage  the  no-load  voltages  EQ  (curve  (1))  must  be 
consumed  as  synchronous  impedance  drops  by  the  armature 
currents  (curve  (2)).  The  ordinates  of  curve  (1)  divided  by  the 
corresponding  ordinates  of  curve  (2)  give  the  values  of  the  syn- 
chronous impedance  corresponding  to  the  various  values  of 
exciting  current  (curve  (3)). 

If  these  values  of  Zs  or  xs  are  multiplied  by  full-load  current 
and  subtracted  from  the  ordinates  of  curve  (1),  the  result  is 
curve  (5)  which  may  approximate  very  closely  to  the  full-load, 
zero-power-factor,  saturation  curve.  Using  this  curve  the  regu- 
lation can  be  obtained  as  in  method  (&). 

In  machines  of  low  saturation,  low  reactance  and  small  mag- 
netic leakage,  a  curve  such  as  (4)  obtained  by  moving  curve  (1) 
to  the  right  through  a  constant  distance  OB  will  be  fairly  close 
to  the  required  zero-power-factor  saturation  curve,  but  in  ma- 
chines of  high  saturation  large  reactance  and  large  magnetic 
leakage  the  curve  obtained  in  this  way  is  much  too  high.  This 
latter  is  the  method  suggested  by  the  standards  committee. 

In  Fig.  310  the  broken  line  shows  the  upper  part  of  a  no-load 
saturation  curve  calculated  using  the  full-load  leakage  factor. 
When  this  correction  is  made  the  horizontal  distance  between  the 
no-load  and  the  full-load  zero-power-factor  saturation  curves  is 
more  nearly  constant  but  is  still  considerably  increased  on  the 
upper  part  of  the  curve. 

The  triangle  pqr,  Fig.  310,  has  as  its  horizontal  side  the  line  pq 
which  represents  the  effect  of  armature  reaction  expressed  in 
ampere-turns  and  as  its  vertical  side  the  line  qr  representing 
a  voltage  drop,  which  is  the  true  reactance  drop  Ix.  The  distance 
between  the  two  curves  measured  in  the  direction  pr  should  be 
very  nearly  constant. 

262.  Regulation. — The  voltage  regulation  of  an  alternator  is 
expressed  as  the  per  cent,  rise  of  the  terminal  voltage  when  full 
load  is  removed  without  changing  the  speed  or  field  excitation. 
The  regulation  is  usually  expressed  for  100  per  cent,  power-factor 
and  80  per  cent,  power-factor  loads.  Usual  values  are  6  to  9  per 
cent,  at  100  per  cent,  power  factor  and  15  to  25  per  cent,  for  80 

per  cent,  power  factor.     Where  automatic  voltage  regulators  are 
21 


322  ELECTRICAL  ENGINEERING 

used  to  maintain  constant  terminal  voltage  close  regulation  is  not 
necessary.  In  modern  machines  of  large  output  values  of 
regulation  as  low  as  14  or  15  per  cent,  at  100  per  cent,  power 
factor  are  considered  satisfactory.  Such  poor  regulation  in- 
dicates a  large  synchronous  reactance  which  limits  the  short- 
circuit  current  to  one  and  one-half  or  two  times  the  full-load 
current  and  the  instantaneous  short-circuit  current  to  six  or  eight 
times  full-load  current.  This  minimizes  the  stresses  on  the  end 
turns  which  are  proportional  to  the  square  of  the  current.  In 
many  cases  reactance  coils  are  connected  in  series  with  alternators 
for  the  same  purpose. 

The  voltage  regulation  depends  mainly  on  two  factors:  (a) 
the  ratio  of  field  strength  to  armature  strength,  and  (6)  the  degree 
of  saturation  of  the  iron  parts  of  the  magnetic  circuit. 

The  excitation  regulation  of  an  alternator  is  the  per  cent. 
increase  of  field  excitation  required  to  maintain  constant  terminal 
voltage  as  the  load  increases  from  zero  to  full  load.  In  Fig.  310 

the  excitation  regulation  at  100  per  cent,  power  factjr  is  - 

100  per  cent.  =  14  per  cent,  and  at  80  per  cent,  power  factor  it  is 

o>c 

-  X  100  per  cent.  =  43  per  cent. 
oo> 

The  extra  excitation  has  three  components: 

1.  That  required  to  counteract  the  effect  of  armature  reaction 
especially  its  demagnetizing  component. 

2.  That  required  to  make  up  for  the  voltage  drop  due  to  the 
reactance  and  resistance  of  the  armature  winding. 

3.  That  required  to  provide  for  the  increase  in  the  magnetic 
leakage. 

The  excitation  regulation  of  an  alternator  under  different  con- 
ditions of  load  and  power  factor  may  be  obtained  from  the  com- 
pounding curves,  Fig.  309,  or  from  to  no-load  and  full-load  satura- 
tion curves,  Fig.  310. 

(a)  Fig.  315  (a)  shows  four  no-load  saturation  curves  drawn  for 
a  machine  with  its  magnetic  circuit  unsaturated  but  with  the 
length  of  the  air  gap  increased  by  equal  amounts  from  100  per 
cent,  in  (1)  to  400  per  cent,  in  (4),  the  armature  strength  remain- 
ing constant.  The  field  strength  increases  directly  as  the  air-gap 

length  and  the  regulation  as  represented  by  ^^>    -~A  -        and 


604 

is  very  much  improved.     In  alternators  of  ordinary  design 


SYNCHRONOUS  MACHINERY 


323 


constants  the  field  strength  is  two  to  three  times  the  armature 
strength  and  the  air  gaps  are  much  longer  than  in  direct-current 
generators. 

Fig.  315(6)  shows  four  saturation  curves  for  a  machine  with  a 
fixed  gap  but  with  increasing  saturation  of  the  iron  parts  of  the 
magnetic  circuit.  The  voltage  regulation  is  very  much  improved 
by  saturation. 

Such  machines  require  a  minimum  of  iron  in  the  magnetic 
circuit  and  are  light  in  weight  but  it  is  difficult  to  obtain  any 
considerable  increase  of  voltage  without  overheating  the  fields. 


1         2        3 


Field  Excitation 


Field  Excitation  u          Field  Excitation 

Effect  of  Saturation  on  Regulation 


Effect  of  Gap  Length  on  Regulation 

(a)  (b)  to 

FIG.  315. — Regulation  of  alternators. 

In  Fig.  315(c),  Oci  and  Ocs  are  saturation  curves  for  two 
machines  with  low  densities  in  the  iron;  the  gap  in  the  second  one 
is  twice  the  length  of  that  in  the  first.  Oc2  and  Oc4  are  curves 
for  similar  machines  with  saturated  magnetic  circuits.  The 
improvement  in  the  regulation  due  to  saturation  is  not  so  great 
in  the  machine  with  the  long  gap  as  in  the  other. 

263.  Short-circuit  of  Alternators. — If  the  impedance  of  the 
load  circuit  of  an  alternator,  with  normal  excitation,  is  gradually 
reduced  to  zero,  the  short-circuit  current  is  limited  only  by  the 
synchronous  impedance  of  the  armature;  it  is 


E 


Zs 


(294) 


and  ranges  from  one  and  one-half  to  three  or  four  times  full-load 
current,  the  larger  values  occurring  in  machines  of  low  synchron- 
ous reactance  and  good  regulation. 

In  the  case  of  sudden  short-circuits  the  momentary  current 

ET 

will  be  much  larger  than  the  permanent  value  -~r»  because  the 

&8 

component  x'  of  the  synchronous  reactance,  which  represents 


324 


ELECTRICAL  ENGINEERING 


the  effect  of  armature  reaction,  does  not  act  instantaneously  to 
limit  the  current.  It  represents  a  change  in  the  flux  which  inter- 
links with  the  field  circuit  of  the  machine  and  on  account  of  the 
inductance  of  the  field  winding  with  its  large  number  of  turns, 
this  change  of  flux  cannot  take  place  instantaneously  but  may 
take  several  seconds  to  become  complete.  The  initial  wave  of 
current  is  limited  only  by  the  true  impedance  of  the  armature 
and  is 

E*  (295) 


The  amplitude  gradually  decreases  as  the  armature  reaction 
reduces  the  flux  in  the  field  and  the  generated  e.m.f.  EQ.  With 
very  large  armature  currents  the  path  of  the  armature  reactance 
flux  becomes  saturated  and  x  is  reduced  below  its  normal  value. 


Armature  Current  in  Phase  1 
"Instantaneous  Maximum 
,  Short  Circuit  Current    Approximately    2^/2  1 sc 

1  Transient  Field  Current 


(a) 


Phase  2 


Armature  Current  in  Phase  2 
.  Pulsating  Field  Current  6  ) 


Normal  Field 
Current 


Normal  Field  Current 
Field  Current 
C*) 
FIG.  316.  —  Short-circuit  currents  in  a  two-phase  alternator. 


J  •' 

I'sc  =  —  7  —  =  —  will  represent  the  short-circuit  current 

\/r2  -j-  x2        x 

only  when  the  short-circuit  occurs  at  the  maximum  point  of  the 
generated  e.m.f.  wave  as  illustrated  in  Fig.  316(a).  Neglecting 
the  armature  resistance  the  generated  e.m.f.  is  consumed  by 

the  back  e.m.f.  of  armature  inductance  =  —  L  ~v'  The  cross- 
hatched  area  under  the  e.m.f.  wave  is  I  edt  and  is  =  I  —L-j-dt 

JO  JV2I'.<  di 

=  \/2I'toL  and  it  causes  the  current  to  increase  from  0  to 


SYNCHRONOUS  MACHINERY        325 

•V/2/'«c-  The  succeeding  quarter  wave  of  e.m.f.  is  negative  and 
reduces  the  current  to  zero  again;  it  then  reverses  and  grows  in 
the  opposite  direction  to  a  value  slightly  less  than  before  since  the 
armature  reaction  will  have  caused  a  slight  reduction  in  the  flux 
and  the  generated  e.m.f.  The  current  is  practically  symmetrical 
about  the  zero  line  and  its  amplitude  decreases  until  the  reduction 
of  flux  by  armature  reaction  is  complete  after  which  it  remains 
constant.  The  transition  from  the  initial  to  the  permanent  value 
will  take  from  a  few  hundred  to  thousands  of  cycles  depending  on 
the  inductance  of  the  field  winding.  In  Fig.  316  it  is  represented 
as  covering  only  a  few  cycles. 

Fig.  316(6)  shows  the  short-circuit  current  in  the  second  phase 
of  the  machine.  The  short-circuit  occurs  at  the  zero  point  of 
the  generated  e.m.f.  wave  and  the  positive  area  under  the  wave 
is  twice  as  great  as  before  and  builds  the  current  up  to  double  the 
value  in  (a).  The  current  then  decreases  and  it  alternates  not 
about  the  line  of  zero  current  as  before  but  about  a  line  starting 
above  the  zero  line  by  an  amount  \/218C  on  the  current  scale  and 
sloping  down  to  coincide  with  the  zero  line  after  a  few  cycles. 

The  armature  current  may  be  considered  as  being  composed 
of  a  transient  direct  current,  shown  as  a  broken  line,  and,  super- 
imposed on  it,  the  alternating  current  l'8C  with  its  decreasing 
amplitude. 

The  m.m.f.  of  the  transient  armature  current,  being  fixed  in 
position  on  the  armature,  revolves  relative  to  the  field  poles  and 
produces  a  single-frequency  pulsation  of  the  field  flux  which  dies 
out  with  the  passing  of  the  transient  current.  This  pulsation  of 
the  field  causes  the  generated  e.m.f.  to  become  unsymmetrical 
about  the  zero  line  as  shown  and  so  after  the  first  cycle  the  arma- 
ture current  crosses  the  zero  line  and  after  a  few  cycles  becomes 
symmetrical. 

The  magnitude  of  the  transient  armature  current  depends  on 
the  point  of  the  voltage  wave  at  which  the  short-circuit  takes 
place  and  therefore  the  short-circuit  currents  in  the  various 
phases  of  a  polyphase  alternator  will  not  be  identical. 

In  machines  of  high  self-inductance  and  low  armature  reac- 

/'  /£    _1_     yf 

tion  the  ratio  y^  =  will  be  small  but  in  large  low-fre- 

J-sc  ^ 

quency  turbo-alternators  with  low  self-inductance  and  high  arma- 
ture' reaction  the  ratio  may  be  very  large  In  extreme  cases  the 
instantaneous  short-circuit  current  may  reach  twenty  to  thirty 


326  ELECTRICAL  ENGINEERING 

times  full-load  current.  Machines  should  be  designed  so  that 
the  instantaneous  short-circuit  current  will  not  exceed  five  to 
eight  times  full-load  current  or  if  this  is  not  possible  external 
reactance  must  be  added  to  limit  the  currents. 

Circuit  breakers  are  provided  with  time  limits  so  that  they  will 
not  open  the  circuit  until  the  current  has  fallen  to  its  steady 
value. 

The  pulsation  of  the  field  flux  due  to  the  m.m.f.  of  the  tran- 
sient armature  current  induces  in  the  field  winding  an  alternat- 
ing current  of  normal  frequency  which  opposes  the  pulsation  of 
the  flux;  this  pulsating  field  current  is  superimposed  on  the  nor- 
mal field  current  giving  the  resultant  current  shown  in  Fig.  316(c). 
In  some  cases  the  maximum  field  current  may  be  ten  to  fifteen 
times  its  normal  value. 

If  the  field  poles  are  solid  or  if  they  are  provided  with  damper 
windings,  the  alternating  current  instead  of  appearing  in  the  field 
winding  may  appear  as  eddy  currents  in  the  pole  faces  and  dam- 
pers and  the  field  current  will  scarcely  be  affected  at  all. 


ted  Voltage 

7\ 

u 

u 

u 

*                                                            Current 

1    A    A    A     A     A     // 

V  \ 

11 

V 

V 

V 

o 

V 

V 

V 

V 

V 

Field  Current  with  Double  Frequency  Pulsation 


(6) 

FIG.  317.  —  Single-phase  short  circuit  currents. 

Fig.  3  17  (a)  shows  the  short-circuit  current  of  a  single-phase 
alternator  in  which  the  circuit  was  closed  at  the  maximum  point 
of  the  voltage  wave.  There  is  no  transient  term  and  the  current 
falls  off  from  the  large  initial  value  to  the  permanent  value  as  in 
the  case  of  the  polyphase  machine.  The  current  wave  has, 
however,  a  decided  peak  showing  the  presence  of  a  third  harmonic 
due  to  the  double-frequency  pulsation  of  the  field  current  as 
shown  in  Fig.  317(fe).  This  phenomenon  was  discussed  in  Art. 
255.  If  the  circuit  is  closed  at  any  other  point  on  the  voltage 
wave  the  current  will  have  a  transient  term  and  the  field  current 
will  have  a  single-frequency  pulsation  in  addition  to  the  double 
frequency  pulsation;  the  single-frequency  pulsation  disappears 
after  a  few  cycles  while  the  double-frequency  pulsation  decreases 


SYNCHRONOUS  MACHINERY 


327 


from  a  maximum  at  the  first  instant  to  its  permanent  value. 
Single-phase  generators  should  always  be  provided  with  dampers 
which  reduce  the  pulsation  of  the  field  current  and  the  resulting 
third  harmonic. 

A  short-circuit  of  one  phase  of  a  polyphase  alternator  is  similar 
in  its  effects  to  a  short-circuit  of  a  single-phase  alternator  but  in 
addition  a  third  harmonic  appears  in  the  voltages  of  the  phases 
not  short-circuited. 

264.  Synchronous  Motor. — A  synchronous  motor  is  similar  to 
an  alternator  in  construction  and  may  be  either  single-phase  or 
polyphase.  The  single-phase  motor  is  not  self-starting  and  must 


Direction  of  JJevolving  Field  and  Torque 

Current  :i;i  Phase  1 


t  in  Phi 


(c)  (d) 

FIG.  318. — Two-pole,    two-phase    synchronous    motor. 

be  brought  up  to  synchronous  speed  before  being  connected 
to  the  supply.  It  is,  therefore,  not  used  except  in  special  cases. 
The  polyphase  motor  when  connected  to  the  supply  will  acceler- 
ate and  run  up  to  synchronous  speed  but  only  a  low  voltage 
should  be  impressed  on  it  at  start  or  very  large  lagging  currents 
will  be  drawn  from  the  supply  lines.  In  many  cases,  however, 
polyphase  synchronous  motors  are  started  by  auxiliary  motors 
to  prevent  voltage  fluctuations  on  the  line. 
.  Fig.  3 18  (a)  represents  a  two-phase,  two-pole  motor.  The 


328  ELECTRICAL  ENGINEERING 

armature  is  stationary  and  is  supplied  with  two-phase  alternating 
currents,  Fig.  318(6).  The  armature  m.m.f.  is  constant  in  value 
as  in  the  alternator  and  revolves  at  synchronous  speed  in  the  anti- 
clockwise direction  and  produces  a  revolving  field  of  constant 
value.  Figs.  3 18 (a),  (c)  and  (d)  represent  the  armature  m.m.f.  at 
the  instants  (1),  (2)  and  (3). 

The  speed  of  the  field  is  directly  proportional  to  the  frequency 
of  the  impressed  e.m.f.  and  inversely  proportional  to  the  number 
of  pairs  of  poles;  it  is 

n  =  ~  =  —  rev.  per  sec.  (296) 

P        P 


This  is  the  speed  at  which  the  motor  operates  and  it  is  constant 
independent  of  the  impressed  e.m.f.  of  the  field  excitation  and  of 
the  load. 

265.  Vector  Diagrams  for  a  Synchronous  Motor. — If  an  alter- 
nating e.m.f.  E  is  impressed  on  the  terminals  of  a  synchronous 
motor  of  resistance  r  and  synchronous  reactance  xs  and  a  current  I 
flows  in  the  armature,  the  phase  relation  of  the  current  and  the 
impressed  e.m.f.  depends  on  the  field  excitation. 


M/ 

Power  Factor -100  Parcent  Power  Factor -50  Percent  Lagging 

'M/  0 "  0  0  _  oO°Lag 

FIG.  319.  FIG.  320. 

In  Fig.  319: 

E  =  impressed  e.m.f.,  which  remains  constant. 
7  =  armature  current,  in  this  case,  in  phase  with  E. 
Ir  =  component  of  the  impressed  e.m.f.  consumed  by  the 

resistance,  in  phase  with  the  current. 
Ix8  =  component  of  the  impressed  e.m.f.  consumed  by 

the  reactance  xsj  in  quadrature  to  the  current. 
IZa  =  component  of  the  impressed  e.m.f.  consumed  by 
the  synchronous  impedance,  JZa  =  /(r  +  jxt). 


SYNCHRONOUS  MACHINERY        329 

gfQ  =  component  of  the  impressed  e.m.f.  consumed 
by  the  counter  e.m.f.  of  the  motor. 

EQ  =  counter  e.m.f.  of  the  motor,  generated  in  the 
armature  by  cutting  the  flux  produced  by  the  field 
m.m.f. 

Mf  =  field  m.m.f.  in  quadrature  ahead  of  the  generated 
voltage  EQ.  This  is  the  value  of  field  excitation 
required  to  make  the  power  factor  of  the  motor 
unity  under  the  assumed  conditions  of  load. 

Fig.  320  is  the  vector  diagram  when  the  current  has  the  same 
value  as  before  but  lags  behind  the  impressed  e.m.f.  by  angle 
4>  =  60  degrees. 

Fig.  321  is  the  vector  diagram  when  the  current  leads  by  angle 
<£  =  60  degrees. 


Eo 


Power  Factor=50  Percent  Leading 
0  «  60° Lead 


*, 

FIG.  321. 

Referring  to  these  diagrams  it  is  seen  that  the  field  excitation 
required  in  a  synchronous  motor  to  produce  a  leading  power  factor 
or  to  cause  the  current  to  lead  the  impressed  e.m.f.  is  greater  than 
that  required  to  produce  a  lagging  power  factor  or  to  cause  the 
current  to  lag  behind  the  impressed  e.m.f. 

If,  therefore,  the  field  current  of  a  synchronous  motor  is  varied, 
there  is  no  change  in  speed  as  in  the  direct-current  motor,  but  the 
generated  e.m.f.  EQ  changes  both  its  value  and  its  phase  relation 
with  the  impressed  e.m.f.  E  and  allows  leading  or  lagging  currents 
to  flow  to  make  up  for  the  change  in  excitation;  when  the  field 
current  is  decreased  a  component  of  current  90  degrees  behind  the 
impressed  e.m.f.  flows  in  the  armature  and  magnetizes  the  field 


330  ELECTRICAL  ENGINEERING 

and  when  the  field  current  is  increased  a  component  of  current 
90  degrees  ahead  of  the  e.m.f.  flows  and  demagnetizes  the  field. 

266.  Characteristic  Curves. — The  most  important  character- 
istic curves  of  the  synchronous  motor  are  (1)  the  compounding 
curves,  or  the  relation  between  field  current  and  armature  current 
for  given  values  of  power  factor,   (2)  the  load  characteristics 
showing  the  relation  between  armature  current  and  output  and 
power  factor  and  output  for  a  given  value  of  field  excitation,  and 
(3)  the  phase  characteristics  or  V  curves  showing  the  relation 
between  armature  current  and  field  current  for  given  values  of 
motor  output. 

To  predetermine  these  curves  it  is  necessary  to  know  the  re- 
sistance r  and  synchronous  reactance  xs  of  the  armature  and  to 
have  the  no-load  saturation  curve  showing  the  relation  be- 
tween the  generated  voltage  EQ  and  the  field  current  //  or  field 
m.m.f.M/. 

The  impressed  e.m.f.  E  is  constant. 

267.  Compounding    Curves. — The    voltage    equation    of    the 
synchronous  motor  is 

E    =  E'0  +  IZ8  (297) 

or  E'0  =  E    -  IZa  (298) 

If  the  current  /  is  taken  as  the  line  of  reference,  the  impressed 
e.m.f.  can  be  expressed  in  rectangular  coordinates  as 

E  =  E  cos  0  +  jE  sin  0, 
where  0  is  the  angle  of  lag  of  the  current. 
The  impedance  drop  in  the  armature  is 

Ei  =  I?s  =  Ir  +  jlx. 
The  e.m.f.  consumed  by  the  generated  e.m.f.  is,  therefore, 

E'0  =  (E  cos  0  -  7r)  +  j(E  sin  0  -  Ix.)  (299) 

and  its  absolute  value  is 


E'0=  V(E  cos  0  -  Ir)2  +  (E  sin  0  -  7zfi)2          (300) 
This  relation  can  also  be  obtained  by  reference  to  the  vector 
diagram  in  Fig.  322. 


but  ,          ,        T-,  T 

oa  =  ob  —  ab  =  E  cos  0  —  Ir, 

and  ad  =  bf  -  cf  =  E  sin  0  -  Ix,. 

therefore, 


cos  0  -  Ir)2  +  (E  sin  0  -  Ix8Y 


SYNCHRONOUS  MACHINERY        331 

Fig.  327  represents  the  compounding  curves  for  unity  power 
factor,  80  per  cent,  power  factor  leading  and  80  per  cent,  lagging. 

To  predetermine  these  curves  the  impressed  e.m.f.  E  is  main- 
tained constant,  a  definite  value  of  power  factor  is  chosen  for 
each  curve,  the  armature  current  /  is 
varied  and  the  values  of  E'Q  are  calcu- 
lated   from    equation    (300).       The 
values  of  field  current  //  correspond- 
ing to   the   calculated   values   of  E'o 
are    obtained    from    the    saturation 
curve,   Fig.  325,   and  are  plotted  as 
ordinates. 

268.  Load  Characteristics. — The  power  input  to  the  motor 
armature  is  the  product  of  the  current  and  the  in-phase  com- 
ponent of  impressed  e.m.f.;  it  is 

Pi  =  El  cos  0.  (301) 

The  electrical  power  transformed  into  mechanical  power  is  the 
product  of  the  current  and  the  in-phase  component  of  the 
e.m.f.  E'o,  consumed  by  the  generated  e.m.f.;  it  is 

P  =  I(E  cos  0  -  Ir)  =  El  cos  0  -  72r,  (302) 

and  is  less  than  the  power  input  by  the  armature  copper  loss. 

The  power  output  is  less  than  the  mechanical  power  developed 
by  the  amount  of  the  constant  losses  in  the  motor,  namely,  the 
iron,  friction  and  windage  losses;  the  output,  therefore,  is 

PZ  =  P  —  constant  losses 

=  PI  —  I2r  —  constant  losses 

=  El  cos  </>  —  I2r  —  constant  losses.  (303) 

Fig.  328  represents  the  load  characteristics  for  a  given  value  of 
field  current  //.  The  value  of  //  chosen  here  is  such  that  the 
back  e.m.f.  of  the  motor  is  greater  than  the  impressed  e.m.f. 
and  the  motor  is  over  excited  at  light  load.  If  a  lower  value  of 
//  had  been  chosen  the  power  factor  would  never  have  reached 
100  per  cent,  and  would  have  been  lagging  through  the  whole 
range  of  load.  Since  //  is  constant  E'Q  is  constant.  /  is  varied 
and  the  corresponding  values  of  cos  <j>  are  obtained  from  equation 
300.  These  values  are  substituted  in  equation  303  and  the 
values  of  /  and  cos  <£  are  plotted  on  a  base  of  power  output. 

At  light  loads  the  power  factor  is  low  and  leading  and  the  motor 


332 


ELECTRICAL  ENGINEERING 


is  over-excited ;  as  the  load  is  increased  the  power  factor  increases 
until  it  reaches  100  per  cent,  at  a  value  of  load  depending  on  the 
field  excitation;  beyond  this  point  the  power  factor  decreases 
again  and  becomes  lagging.  The  current  increases  continually 
but  is  finally  limited  by  the  synchronous  impedance  of  the 
armature. 


FIG.  323. 

These  results  may  be  verified  by  referring  to  the  diagrams  in 
Fig.  323.  In  (2)  the  current  /  is  in  phase  with  the  impressed 
e.m.f .  E,  that  is,  the  power  factor  is  100  per  cent,  and  <f>  =  0.  In 
(3)  the  power  output  is  increased  and  the  current  is  assumed  to 
have  increased  to  double  the  value  in  (2) ;  the  synchronous  imped- 
ance drop  is  also  doubled ;  E'o  remains  constant  in  all  cases  and  its 
position  can  be  found  by  drawing  arcs  of  two  circles  of  radii  E'Q 
and  IZa  about  the  two  ends  of  the  vector  E.  The  synchronous 
impedance  drop  can  be  resolved  into  its  two  components  Ir  and 

Ixa  by  setting  off  the  constant  angle  a  =  tan"1  — .     The  current 

X8 

I  is  then  drawn  in  the  direction  of  Ir  and  it  lags  behind  E  by  the 


SYNCHRONOUS  MACHINERY  333 

angle  0.  The  power  factor  is  cos  <t>  and  is  lagging.  The  power 
output,  neglecting  the  constant  losses,  is  equal  to  the  product  of 
the  current  and  the  in-phase  component  of  the  e.m.f.,  E'o,  con- 
sumed by  the  generated  voltage  EQ  or  it  is  equal  to  the  product 
of  the  e.m.f.,  E'o,  and  the  component  of  the  current  in  phase 
with  it,  and  since  E'o  is  constant  the  power  output  may  be 
represented  by  the  in-phase  component,  of  the  current  OP  in,, 
each  of  the  diagrams. 

In  (1)  the  power  is  decreased  and  the  current  is  taken  as  half 
of  that  in  (2);  the  position  of  E'ois  found  as  before  and  the  current 
leads  E  by  the  angle  $. 

The  power  output  cannot  increase  indefinitely  but  reaches  its 
maximum  when  the  decrease  in  power  factor  overcomes  the  in- 
crease in  current  or  when  the  component  of  the  current  in  phase 
with  the  generated  e.m.f.  begins  to  decrease;  the  motor  then  be- 
comes unstable  and  falls  out  of  synchronism  and  stops.  In  (4) 
the  current  is  increased  to  three  times  the  value  in  (2)  its  compo- 
nent OP  in  phase  with  E'Q  is  greater  than  in  (3)  and  therefore  the 
power  output  is  still  increasing.  In  (5)  the  current  is  four  times 
its  value  in  (2)  but  the  component  OP  in  phase  withE'ois  less  than 
in  (4)  and  the  output  is  therefore  decreasing.  The  position  of 
maximum  output  lies  between  (4)  and  (5). 

The  maximum  power  output  of  a  synchronous  motor  is  far 
beyond  the  limits  set  by  temperature  rise. 

269.  Ph^se  Characteristics. — If  the  field  excitation  of  a  motor 
with  constant  output  is  varied,  the  armature  current  changes 
both  its  value  and  its  phase  relation  with  the  impressed  e.m.f. 
For  each  output  there  is  a  certain  value  of  field  excitation  which 
makes  the  current  a  minimum  and  brings  it  in  phase  with  the 
e.m.f. ;  as  the  excitation  is  decreased  below  this  value  the  current 
increases  and  becomes  lagging;  as  the  excitation  is  increased  the 
current  increases  and  becomes  leading. 

In  Fig.  329  (a)  are  shown  the  phase  characteristics  for  outputs, 
PI  =  0  or  at  no  load,  P2  =  full  load  and  P2  =  twice  full  load. 

For  each  curve  the  output  Pz  =  El  cos  <f>  —  I2r  —  constant 
losses  is  kept  constant;  thus, 

PZ  +  I2r  +  constant  losses 


cos  0  = 


EI 


E'Q  =  V(E  cos  <f>  -  7r)2  +  (E  sin  0  -  Ix8)2. 

As  /  varies  the  corresponding  value  of  cos  <£  is  found  from  equa- 
tion (304)  and  by  substituting  /,  cos  <£  and  sin  </>  in  equation  (300), 


334  ELECTRICAL  ENGINEERING 

the  values  of  E'Q  are  found.  These  are  replaced  by  the  corre- 
sponding values  of  field  current  //  obtained  from  the  no-load 
saturation  curve. 

The  lowest  point  on  each  curve  represents  the  smallest  current 
input  for  the  given  output  and  thus  represents  the  condition  of 
unity  power  factor.  The  curve  joining  these  lowest  points  is  the 
compounding  curve  for  unity  power  factor.  If  the  phase  char- 
acteristics are  very  steep  a  slight  change  in  field  excitation  pro- 
duces a  large  change  in  armature  current,  or  a  large  component 
of  wattless  current  is  required  to  correct  for  a  slight  variation  in 
field  excitation.  This  is  the  case  in  a  motor  with  small  synchro- 
nous reactance  or  small  armature  reaction  and  the  motor  is  un- 
stable. If  the  synchronous  reactance  is  large  only  a  slight  change 
in  armature  current  is  produced  by  a  change  in  field  excitation 
and  the  phase  characteristics  are  flat  and  the  motor  is  stable. 

270.  Torque.  —  Since  the  speed  of  the  synchronous  motor  is 
constant,  the  torque  developed  at  any  output  is  directly  propor- 
tional to  the  output. 

If  T  is  the  torque  in  pounds  at  1-ft.  radius  developed  in  the  ar- 
mature, the  electrical  power  transformed  into  mechanical  power  is 


X  746  Watts  =  EI  COS  *  ~ 


and  thus  the  torque  developed  is 


and  the  torque  available  for  the  load  is 

m        EI  cos  <£  —  Pr  —  constant  losses  xx  33,000  1U    .        /onAs 
27T  X  r.p.m.  ^W 

271.  Blondel  Diagram  for  a  Synchronous  Motor.  —  In  Fig.  324 
the  triangle  OAB  is  a  voltage  diagram  for  a  synchronous  motor. 
OA  =  E  is  the  constant  impressed  voltage,  AB  =  E0  is  the  back 
voltage  of  the  motor  and  OB  =  E'  =  IZ8  is  the  synchronous 
impedance  drop  due  to  current  7.  The  angle  OAB  =  a  is  called 
the  coupling  angle. 

The  current  I  lags  behind  E  by  angle  4>  and  behind  E'  by  angle 


0  =  tan 

OC  and  AC  are  two  straight  lines  making  equal  angles  6  with 
the  ends  of  the  constant  impressed  e.m.f.  vector  OA. 


SYNCHRONOUS  MACHINERY 


335 


The  current  may  be  represented  by  the  vector  OB  if  its  scale 
is  Za  times  the  voltage  scale  and  the  phase  relation  of  the  current 
and  the  impressed  e.m.f.  is  found  as  the  inclination  of  OB  to  OC 
which  is  equal  to  the  phase  angle  <£.  The  component  of  the 
current  in  phase  with  the  impressed  e.m.f.  is  Oa,  the  projection 
of  OB  on  OC. ' 

For  constant  power  input  to  the  motor  this  in-phase  current 
Oa  must  remain  constant  and  the  locus  of  B  is  therefore  a  straight 
line  through  a  perpendicular  to  OC. 


E ^  A          0  E  A 

©  © 

FIG.  324. — Blondel  diagram. 

The  current  Oa  may  be  expressed  in  terms  of  E  and  Eo  by 
dropping  a  perpendicular  from  A  on  OC  at  6,  Fig.  324  (2),  remem- 
bering that  voltage  values  are  changed  to  current  values  by 
dividing  by  Zs. 

77?  77? 

Oa   =  Ob  +  ba  =  ^  cos  0+-^  cos  {180  -  (0  +  a)} 

77?  77? 

=  -^  cos  0  —  -^  cos  (B  +  a) 


and  the  power  input  is 

77*2 
Pi  =  E  X  Oa  =  ~  cos  0  - 


cos  (6  +  a) 


(307) 


A  similar  expression  for  the  power  output  may  be  obtained  by 
drawing  the  line  cBd  making  angle  9  with  AB,  Fig.  324(3)  and 
dropping  perpendiculars  from  0  and  A  on  it. 


336  ELECTRICAL  ENGINEERING 

The  component  of  current  in  phase  with  the  generated  voltage 
is 

cB  =  cd  —  Bd  =  Y  cos  (0  —  <*)  —  -~°  cos  0 
and  the  power  output  is 

777  77T  Try     9 

P2  =  E0  X  cB  =  -^  cos  (0  -  a)  -  ~-  cos  0.        (308) 

It  is  now  required  to  find  the  equation  of  the  locus  of  B  which 
will  make  P2  constant  as  the  excitation  and  therefore  the  gener- 
ated voltage  EQ  varies  and  the  current  varies. 

Take  two  rectangular  axes,  Fig.  324(4),  the  X-axis  along  AC 
and  the  F-axis  at  right  angles  to  it,  then 

x2  +  2/2  =  #o2  (309) 

and 

x  =  E0  cos  (0  -  a).  (310) 

Substituting  these  in  (308) 


D  (  ->   i      9\  cos 

Pi   =   -yT    ^-  (Z2   +    2/2)  -0T 

63  ^« 

and 


P2r 


cos  0  cos  0  cos20 

and  completing  squares 

-7-  ~  P& 


(SID 

v       ; 


.. 

2  cos  6*7  4  cos2  0       cos2  0          cos2  0 

this  is  the  equation  of  the  locus  of  B  and  it  is  a  circle  of  radius 


R  -  -  =     u         -  P2r  (312) 


-  . 

and  having  as  its  center  the  point  f  ^  -  z'  Oj  »  which  is  the  point 

C. 

Thus  the  lines*  of  constant  power  output  P%  are  circles  of  radius 

R  =  —  \/^  -  P2r  drawn  about  the  point  C,  Fig.  324(5).     The 
r  \  4 

circle  corresponding  to  P2  =  0  passes  through  0  and  A. 

The  maximum  possible  power  output  would  correspond  to  the 


SYNCHRONOUS  MACHINERY  337 

case  where  the  radius  =  0  and  is  represented  by  the  point  C. 
This  maximum  power  P2  is  found  by  equating  the  equation  for 
the  radius  to  zero 


and  therefore 


P2  =  f?  (313) 


This  is  the  maximum  power  output  with  maximum  excitation 
and  is  far  beyond  the  practical  limit  of  output. 

The  maximum  output  with  a  given  excitation  I/  corresponding 
to  a  generated  voltage  EQ  may  be  found  by  differentiating  P2 
with  respect  to  a, 

777  77F  "Jjt    2 

P2  =  — =-2  cos  (6  —  a) ^?-cos  B 

63  *   AS 

sin  (0  —  a)  =  0,  for  maximum;         (314) 


da         Z8 

the  condition  of  maximum  output  is 

sin  (B  —  a)  =  0,  or  a  =  6 
and  substituting  this  in  the  equation  for  P2, 

-  ^cos  0  (315) 


This  corresponds  to  the  condition  when  B  falls  on  AC.  For 
positions  of  B  to  the  right  of  AC  the  power  decreases  again.  If  a 
motor  is  loaded  until  the  coupling  angle  a.  becomes  greater  than 
6  it  will  fall  out  of  step  and  stop.  The  line  AC,  therefore,  marks 
the  limit  of  stable  operation. 

When  the  point  B  falls  on  the  line  OC  the  current  is  in  phase 
with  the  impressed  voltage  and  the  power  factor  is  unity.  This  is 
the  position  of  minimum  current  for  a  given  output.  For  points 
to  the  right  of  OC  the  current  is  lagging  and  for  points  to  the  left 
the  current  is  leading. 
pi 

The  value  P2  =  T~  may  be  found  by  differentiating 

EE0        E02 

—^-        -fj-  cos  6  with  respect  to  Eo, 


—  =  -=- =—  cos  B  =  0  for  maximum; 

22 


338  ELECTRICAL  ENGINEERING 

E 
therefore,  2Eo  cos  6  =  E  and  EQ  —  r  -  -  and  the  maximum 

2i  COS  u 

value  of  P2max.  is 

E  E       2  cos  e 


p  I  \ 

2  "  Z8  2  cos  0    "  \2  cos  B) 


Z8  2  cos  0       \2  cos  B       Z8 

E2      E2      E2      ,    , 
=  p:  --  7-  =  -7-  as  before. 
2r       4r       4r 

To  find  the  relation  between  xa  and  r  which  makes  P2max.  a 

_  r  r 

maximum,  substitute  Z8  =  v  r2  +  xa2  and  cos  6  =  ~-  = 


/  2  . 
v  T  ~r  £«  » 

assume  r  to  be  constant  and  differentiate  P2  max.  with  respect 
to  xs. 


2max. 


dxe 

=  —  /  2   i   °  l\%  ~t~  (  2  _L — iV2  =  0  f°r  maximum, 

and  assuming  EQ  =  E,  which  would  be  approximately  correct 
for  normal  operation, 

-x.Vr2  +  x,2  +  2rxs  =  0 
and  from  this 

x,  =  V3r. 

This  is  a  much  lower  value  of  x8  than  is  usually  found  in  such 
machines. 

272.  Synchronizing  Power. — By  synchronizing  power  is  meant 
the  change  in  power  per  degree  change  of  the  coupling  angle  a,  it 
represents  the  stiffness  of  the  coupling  of  the  motor  and  w&s 
found  in  equation  (314) 

dPz       EE$         /n         ,. 
_  =  -_  sin  (e.  -  a) ; 
da.          Zj8 

it  is  directly  proportional  to  the  excitation  or  EQ  and  is  inversely 

proportional  to  the  synchronous  impedance  and  therefore  to  the 

ja1  Jf 
armature  strength.     It  varies  from  a  maximum  value  -~—   sin 


6  —     y8  when  a  =  0  to  zero  when  a  =  6.     If  a  becomes  greater 

•"« 
than  6  the  machine  cannot  pull  into  step  again. 


SYNCHRONOUS  MACHINERY 


339 


273.  Construction  of  the  Characterictic  Curves  of  a  Synchro- 
nous Motor  from  the  Blondel  Diagram. — A  three-phase,  150-kw. 
(output)  synchronous  motor  has  a  resistance  per  phase  r  =  0.04 
ohms  and  a  synchronous  reactance  xa  =  0.6  ohms.  The  con- 
stant losses  may  be  taken  as  5  per  cent,  of  the  rated  output  = 
7,500  watts. 


500 
400 
300 
300 
100 
0 

X 

X 

X" 

> 

/ 

/ 

T 

/ 

/ 

/ 

A 

, 

^_ 

' 

5    10    15    20    2? 

Impressed  Voltage  per  Phase 

E=^f  =**  Volts 
Power  Output  per  Phase  at 

Full  Load 

P2= 60  Kilowatts 

Constant  Loss  per  Phaeo 

Pl  =  2.6  Kilowatts 

Resistance  per  Phasa 

=  0.04  Ohms 

cbroiious  Keartanc* 
per  Phase 
0?.=0.6  Ohms 


FIG.  325. 

In  constructing  the  Blondel  diagram  it  is  as  well  to  work  on 
one  phase  only  with  a  full-load  output  of  50  kw.  Assuming  that 
the  armature  is  F-connected,  the  impressed  voltage  per  phase 

440 

is  E  =  — -f=  =  254  volts  and  this  is  maintained  constant.     The 
\/3 

no-load  saturation  curve  is  shown  in  Fig.  325. 


Perdent  P.  F. 

U  Load  I  La* 
Current 


FIG.  326. 

In  Fig.  326  the  horizontal  6 A  =  E  =  254  volts  and  the  two 

1* 

lines  OC  and  AC  are  drawn  making  angles  0  =  tan"1  —  =  tan-1 
15  with  the  ends  of  OA ;  the  point  of  intersection  C  is  not  shown. 


340  ELECTRICAL  ENGINEERING 

The  length  of  each  of  the  two  sides  OC  and  AC  is 

_A_  254 

2  cos  0  0.04      1>yuo- 


The  circles  representing  constant  power  outputs  per  phase 
from  0  to  110  kw.  are  shown.  Their  radii  may  be  found  by 
substituting  the  given  values  of  output  in  the  equation. 

B  =  Z>    ' 

but  here  P%  is  the  power  developed  at  the  shaft  and  is  greater 
than  the  output  by  the  amount  of  the  constant  losses  per  phase 
=  2,500  watts. 
For  example,  the  radius  of  the  circle  for  full-load  output  is 


*  -  v    -  **•  =  o 

=  1,776. 
and  the  radius  for  zero  output  is 


0.60    / 

R  ~  KoiV 


2  -  (2,500)  (0.04)  =  1,899. 


4 

274.  Load  Characteristics. — To  construct  the  load  character- 
istics, take  any  value  of  motor  back  voltage  EQ  and  draw  a  circle 
about  A  with  EQ  as  radius.  Here  EQ  is  taken  as  278  volts  cor- 
responding to  a  field  current  //  =  10  amp.  and  this  gives  unity 
power  factor  at  full  load.  At  the  point  B,  Fig.  326,  AB  represents 
the  constant .  motor  voltage  EQ  =  278,  OB  represents  the  syn- 
chronous impedance  drop  E1  =  IZa  at  full  load  and  it  also  repre- 

129       129 
sents  the  current.     OB  =  129  volts  or  =  -~-  =  7:-^  =  215  amp.; 

Z/  a  U.O 

the  power  factor  =  100  per  cent.     At  the  point  Bi  the  output 
=  90  kw.  per  phase,  the  current  =  — «-^  =  7^-  =  413  amp.  and 

&8  U.O 

the  power  factor  cos  BOBi  =  cos  19.5  degrees  =  0.943  =  94.3 
per  cent,  lagging. 

At  #3  the  output  =  0,  the  current  =  —^  =  ^  =  40  amp. 

/j  a  U.O 

and  the  power  factor  cos  <J>  =  cos  BOB3  =  cos  75  degrees  = 
0.259  =  25.9  per  cent. 

The  maximum  power  output  for  the  given  excitation  occurs  at 
#2  where  the  circle  cuts  the  line  AC.  Here  output  =  106.5  kw. 


SYNCHRONOUS  MACHINERY 


341 


per  phase,  the  current  =  605  amp.  and  the  power  factor  =  79  per 
cent.     To  the  right  of  AC  operation  is  unstable. 

These  values  of  current  and  power  factor  are  plotted  on  an 
output  base  in  Fig.  328. 


0   40   80  120  100  200  240  280  820  800  400 

Amperes  Load  Current 

FIG.  327. — Compounding  curves  of  a 
synchronous  motor. 


0     10      20     80      40     60     «0      70      80     90     100    110 
Kilowatts  Output  per  Phase 

FIG.  328. — Load  characteristics  of  a 
synchronous  motor. 


275.  Phase  Characteristics  or  "V"  Curves  and  Compounding 
Curves.  —  To  obtain,  the  "V"  curve  for  full-load  output  take 

points  as  B,  B±  and  B*>  on  the  50-kw.  circle.     At  B  the  current 
1  2Q 
-TT  =  215  amp.,  the  power  factor  =  cos  <f>  =  1.00,  the 


s 


motor  back-voltage  =  AB  =  278  volts  and  the  corresponding 


5  10  15  20  25          0 

Amperes  Field  Current 


"V"  curves. 


FIG.  329. 


5  10  15  20 

Amperes  Field  Current 

w 

Power-factor  curves. 


field  current  is  10  amp.,  Fig.  325;  at  B±  the  current  is  250  amp., 
the  power  factor  is  cos  30  degrees  =  0.866  =  86.6  per  cent,  lead- 
ing, the  motor  voltage  is  350  volts  and  the  field  current  is  13.8 
amp.;  at  B*,  the  current  is  250  amp.  the  power  factor  is  86.6  per 
cent,  lagging  and  the  field  current  is  7.4  amp.  The  "V"  curve? 


342  ELECTRICAL  ENGINEERING 

for  zero  output,  full  load  and  twice  full  load,  are  shown  in  Fig. 
329 (a).  The  minimum  armature  current  for  each  output  occurs 
where  the  power  circle  cuts  the  line  OC,  that  is,  where  the  power 
factor  is  unity.  The  corresponding  values  of  the  power  factor 
are  plotted  on  a  field  current  base  in  Fig.  329(6). 

Compounding  Curves. — In  Fig.  326(6)  circles  are  drawn  about 
center  O  with  radii  corresponding  to  currents  from  one-half 
to  twice  full-load  current.  Using  the  points  where  these  cut  the 
constant  power-factor  lines  the  compounding  curves  in  Fig.  327 
are  drawn  for  unity  power  factor,  80  per  cent,  power-factor  lead 
and  80  per  cent,  power-factor  lag. 

The  theoretical  maximum  possible  power  output  of  this  motor 

E2         C254D2 
would  be  P  =  -£r  =  4V     Q  Q4  =  403,000  watts  =  403  kw.  per 

phase  which  is  more  than  eight  times  full  load.  Such  an  output 
would  require  a  motor  back  voltage  =  AC  =  1,905  volts. 

276.  Starting  Synchronous  Motors. — The  single-phase  syn- 
chronous motor  is  not  self-starting  and  must  be  brought  up  to 
synchronous  speed  before  being  connected  to  the  supply.  This 
is  due  to  the  fact  that  single-phase  armature  reaction  does  not 
produce  a  revolving  field.  Single-phase  motors  are,  therefore, 
not  used  except  in  special  cases  and  in  small  sizes. 

Polyphase  synchronous  motors  are  inherently  self -starting  and 
when  connected  to  the  supply  will  accelerate  and  run  up  to  syn- 
chronous speed,  but  only  a  low  voltage  should  be  impressed  at 
start  or  very  large  lagging  currents  will  be  drawn  from  the  supply 
lines.  In  order  to  do  away  with  the  large  starting  currents  a 
great  many  polyphase  synchronous  motors  are  started  by  auxil- 
iary motors;  they  must  be  brought  up  to  synchronous  speed  and 
synchronized  just  as  alternators.  If  the  exciter  for  the  motor  is 
mounted  on  the  same  shaft  it  may  be  used  as  a  starting  motor 
or,  if  the  motor  is  one  unit  of  a  motor-generator  set,  the  genera- 
tor may  be  used  as  a  starting  motor,  but  in  both  cases  a  supply 
of  direct  current  is  necessary. 

An  induction  motor  mounted  on  the  same  shaft  may  be  used 
to  start  the  synchronous  motor  but  it  must  have  a  smaller  num- 
ber of  poles  and  therefore  a  higher  synchronous  speed  than  the 
motor  to  be  started.  The  synchronous  motor  is  raised  above 
synchronous  speed,  the  induction  motor  is  then  disconnected, 
and  the  synchronizing  switch  is  closed  as  the  motor  passes  through 
synchronous  speed. 


SYNCHRONOUS  MACHINERY 


343 


277.  Self-starting  Motors. — In  order  to  improve  the  starting 
torque  of  synchronous  motors  and  also  to  prevent  hunting; 
short-circuited  grids  are  placed  in  the  pole  faces  or  between  the 
poles  and  sometimes  complete  squirrel-cage  windings  are  carried 
by  the  rotors  (Fig.  330).  The  field  winding  at  start  may  be 
either  open  or  short-circuited.  In  the  following  discussion  it  will 
be  assumed  that  the  field  circuit  is  open.  When  a  motor  at  rest 
is  connected  to  a  polyphase  supply  torque  is  developed  in  two 
ways:  (1)  the  revolving  field  sweeps  across  the  pole  faces  and 
grids  and  generates  e.m.fs.  and  currents  in  them;  these  currents 


FIG.  330. — Rotor  of  a  synchronous  motor  with  a  squirrel-cage  winding. 

react  on  the  field  and  produce  torque  which  causes  the  rotor  to 
follow  the  field.  The  rotor  can,  however,  never  be  brought  up  to 
synchronous  speed  by  this  torque  because  the  e.m.fs.  and  currents 
are  induced  only  below  synchronous  speed.  The  motor  starts  as 
a  regular  induction  motor  and  when  it  is  nearly  up  to  speed  the 
field  locks  with  the  revolving  armature  in  the  position  of  minimum 
reluctance. 

2.  The  second  method  of  producing  torque  for  starting  depends 
on  hysteresis;  as  the  armature  north  pole  passes  a  field  pole  it 
induces  a  south  pole  in  it  and  the  attraction  between  the  two 
tends  to  make  the  rotor  follow  the  armature  field.  Due  to 
hysteresis  this  induced  field  pole  persists  after  the  armature  pole 
has  passed;  the  succeeding  armature  south  pole  first 'repels  the 
field  pole  then  while  passing  changes  it  to  a  north  pole  and  at- 


344  ELECTRICAL  ENGINEERING 

tracts  it  in  the  direction  of  rotation.  The  resulting  torque  is 
small  at  low  rotor  speeds  but  becomes  greater  as  synchronous 
speed  is  approached  and  the  field  pole  remains  longer  under  the 
influence  of  the  armature  pole.  When  the  armature  pole  is 
moving  very  slowly  across  the  field  pole  the  two  lock  in  the  posi- 
tion of  minimum  reluctance. 

Since  the  grids  or  squirrel-cage  windings  are  carrying  current 
only  while  starting  up,  or  when  there  is  relative  motion  between 
the  armature  reaction  flux  and  the  field  poles,  as  when  there  is 
hunting,  it  is  not  objectionable  if  they  have  a  comparatively  high 
resistance;  in  fact,  the  higher  the  resistance  the  greater  will  be  the 
starting  torque,  within  limits.  The  bars  of  the  squirrel  cage  can 
therefore  be  made  of  any  suitable  material  and  steel  is  often 
employed  but  the  end  rings  are  generally  made  of  brass. 

When  the  motor  is  running  at  synchronous  speed  the  field 
circuit  may  be  closed  and  the  impressed  e.m.f.  raised  to  its  full 
value.  On  closing  the  field  circuit  it  may  be  found  that  the 
polarity  is  wrong  and  the  rotor  will  then  drop  back  in  phase  by 
180  degrees  and  this  change  will  be  accompanied  by  a  sudden 
rush  of  current.  To  avoid  this  it  is  better  to  excite  the  fields 
through  a  large  resistance  just  before  synchronism  is  reached, 
then  increase  the  field  to  normal  and  raise  the  impressed  voltage 
to  full  value.  When  starting  in  this  way  the  motor  draws  a  very 
large  lagging  current  since  the  impressed  voltage  is  consumed  by 
the  synchronous  impedance  of  the  armature,  and  the  power  factor 
is  very  low.  The  impressed  voltage  at  start  must  be  reduced 
to  about  one-third  of  its  full  value  in  order  to  reduce  the  starting 
current.  It  is  the  flux  of  armature  reaction  which  produces  the 
starting  torque  and,  therefore,  a  motor  with  high  armature  reac- 
tion will  give  better  starting  torque  than  one  of  low  armature 
reaction. 

When  the  motor  is  running  below  synchronous  speed  large 
voltages  sometimes  reaching  5,000  volts  or  more  are  induced  in 
the  open  field  winding  by  the  revolving  armature  flux.  Such 
voltages  are  dangerous  and  may  puncture  the  insulation  of  the 
field  or  endanger  the  lives  of  operators.  Both  the  magnitude  and 
frequency  of  these  induced  voltages  becoine  zero  when  the  motor 
reaches  synchronous  speed. 

The  potential  stresses  may  be  reduced  by  breaking  up  the  field 
into  sections,  by  means  of  a  suitable  switch,  during  the  starting 
period. 


SYNCHRONOUS  MACHINERY  345 

The  approach  of  synchronism  may  be  recognized  by  the  sound 
or  an  alternating-current  voltmeter  connected  across  a  section  of 
the  field  winding  will  indicate  synchronism  by  a  zero  reading. 

If  the  field  winding  is  short-circuited  while  coming  up  to  syn- 
chronism these  large  induced  voltages  will  not  exist  but  the  start- 
ing torque  of  the  motor  will  in  general  be  reduced.  The  short- 
circuited  field  winding  tends  to  act  in  somewhat  the  same  way  as 
the  grids  but  its  inductance  is  so  high  that  the  flux  produced 
through  it  by  the  armature  m.m.f .  is  small  and  this  reduction  of 
the  flux  in  the  field  poles  more  than  counterbalances  the  torque 
due  to  induced  currents.  In  the  majority  of  cases,  however,  it  is 
better  to  start  the  motor  with  the  field  winding  short  circuited 
since  the  safety  of  operation  is  thereby  increased.  When  syn- 
chronous speed  is  reached  the  full  excitation  should  be  applied 
and  then  the  impressed  voltage  raised  to  full  value. 

When  individual  exciters  are  used  the  motor  field  may  be 
short-circuited  by  reducing  the  exciter  field  to  its  lowest  value. 

278.  Synchronous  Phase  Modifier. — Since  by  varying  the  field 
excitation  of  a  synchronous  motor  the  power  factor  can  be  made 
either  leading  or  lagging,  such  machines  can  be  used  to  improve 
the  power  factor  of  transmission  lines  or  distributing  circuits  by 
drawing  wattless  leading  currents  to  compensate  for  the  wattless 
lagging  currents  required  by  the  load.  The  fields  must  be  over- 
excited and  the  synchronous  reactance  should  not  be  very  large. 
This  is  one  of  the  most  important  characteristics  of  the  synchro- 
nous motor  and  is  being  applied  to  an  ever-increasing  extent. 

A  synchronous  motor  used  in  this  way  is  called  a  synchronous 
phase  modifier  and  is  usually  operated  without  load  drawing  the 
required  wattless  leading  current  and  the  small  power  current 
supplying  its  own  losses.  In  some  cases,  however,  it  may  be 
advantageous  to  supply  some  load  from  it. 

In  the  case  of  long-distance  transmission  lines  the  synchronous 
phase  modifier  is  used  to  obtain  a  constant  terminal  voltage 
under  all  conditions  of  load  with  a  constant  impressed  voltage  at 
the  generating  station.  Assume  a  transmission  line  with  a  con- 
stant impressed  voltage  of  110,000  volts.  Under  normal  condi- 
tions the  terminal  voltage  at  no  load  rises  to  120,000  due  to  the 
compounding  effect  of  the  line  capacity,  while  at  full  load  80  per 
cent,  power  factor  the  terminal  voltage  falls  to  90,000  volts. 
To  maintain  a  constant  terminal  voltage  of  100,000  a  synchro- 
nous phase  modifier  must  be  installed  in  the  terminal  station; 


346  ELECTRICAL  ENGINEERING 

during  periods  of  light  load  it  is  operated  under  excited  drawing 
a  large  lagging  current  to  pull  down  the  voltage  from  120,000  to 
100,000;  while  under  load  it  is  over-excited  and  draws  a  leading 
current  to  raise  the  terminal  voltage  from  90,000  to  100,000. 

The  constant  voltage  regulation  is  made  automatic  by  install- 
ing a  voltage  regulator  operating  on  the  field  circuit  of  the  phase 
modifier  and  having  its  alternating-current  control  magnet  con- 
nected through  a  potential  transformer  across  one  phase  of  the 
circuit  in  which  the  voltage  is  to  be  maintained  constant. 

279.  Parallel  Operation  of  Alternators. — Before  an  alternator 
is  connected  in  parallel  with  another  machine  which  is  supply- 
ing power,  the  incoming  machine  must  be  adjusted  to  give  the 

same  voltage,  must  have  the  same 
frequency  and  must  be  in  phase. 

The  condition  of  synchronism  may 
be  indicated  by  incandescent  lamps  or 
some  form  of  synchroscope  (Art.  452). 

In  Fig.  331  when  the  incoming 
machine  B  is  in  synchronism  with  A 

FlQ'  331p7rli1iteeiniat0rsin  there  is  no  voltage  across  the  switch 

and  the  lamps  are  dark.  The  two 

machines  are  assumed  to  have  been  adjusted  to  give  equal 
voltages. 

If  the  frequency  of  B  is  lower  or  higher  than  that  of  A,  there 
will  be  a  slow  pulsation  of  the  light  showing  the  difference  between 
the  two  frequencies.  When  the  pulsations  are  very  slow  and  the 
periods  of  darkness  long  the  switch  may  be  closed  and  the  two 
machines  will  operate  in  parallel.  Lamps  are  not  very  satis- 
factory, since  they  do  not  show  whether  the  incoming  machine 
is  running  too  slow  or  too  fast. 

280.  Effect  of  Inequality  of  Terminal  Voltage.  — If  two  alter- 
nators are  operating  at  the  same  frequency  and  are  in  phase  but 
have  not  their  fields  adjusted  to  give  the  same  terminal  voltage, 
a  wattless  current  will  flow  between  the  two  machines  leading  and 
magnetizing  in  the  machine  of  lower  field  excitation  and  lagging 
and  demagnetizing  in  the  machine  of  higher  field  excitation. 

If  EB,  the  voltage  of  B}  is  lower  than  EA,  the  voltage  of  A,  then 
E'  the  difference  between  the  two  will  act  in  the  local  circuit 
through  the  two  armatures  in  series  and  will  produce  a  current 
F  lagging  nearly  90  degrees  behind  El  and  EA  and  leading  EB 
(Fig.  332). 


SYNCHRONOUS  MACHINERY  347 

The  circulating  current  is 


7  +7 

4  A     r  ZB 

where  Z4  and  ZB  are  the  synchronous  impedances  of  the  two 
machines. 

This  current  lowers  the  terminal  voltage  of  A  since  it  is  lagging 
in  A  and  raises  the  terminal  voltage  of  B  since  it  is  leading  in  B 
and  the  two  are  made  equal  to  the  load  voltage  E. 


I 

FIG.  332. 

By  adjusting  the  field  rheostats  the  wattless  circulating  cur- 
rents can  be  eliminated  for  any  load,  but  if  the  two  machines 
have  different  voltage  characteristics,  as  the  load  varies  wattless 
currents  will  circulate  to  correct  for  the  differences  of  excitation. 

With  machines  of  reasonably  high  armature  reaction  the 
wattless  cross  currents  are  small  even  with  large  variations  of 
excitation. 

When  two  similar  alternators  in  parallel  are  supplying  an 
inductive  load,  they  should  operate  at  the  same  power  factor. 
If  one  has  a  lower  excitation  than  the  other  it  will  not  supply 
its  proper  share  of  the  watless  kva.  and  will  operate  at  a  higher 
power  factor  than  the  second  machine. 

281.  Effect  of  Inequality  of  Frequency. — Two  alternators 
operating  in  parallel  must  have  the  same  average  frequency,  but 
one  may  instantaneously  drop  behind  or  run  ahead  of  the  other. 

Alternators  driven  by  water  turbines  or  steam  turbines  or 
electric  motors  will  have  a  constant  angular  velocity  but  when  the 
prime  movers  are  steam  engines  or  gas  engines  the  angular  veloc- 
ity will  pulsate  about  its  average  value  during  each  revolution. 

If  two  machines  in  parallel  are  excited  to  give  the  same  termi- 
nal voltage  and  one  falls  behind  the  other,  a  power  cross-current 
will  circulate  through  the  armatures  and  transfer  energy  from  the 
leading  to  the  lagging  machine. 


348  ELECTRICAL  ENGINEERING 

Fig.  333  shows  the  case  of  two  machines  of  which  A  is  ahead 
of  its  normal  position  by  angle  a  and  B  is  behind  by  angle  a. 
A  is  therefore  ahead  of  B  by  angle  2a  and  the  resultant  voltage 
causing  a  current  to  circulate  through  the  two  armatures  is 

E'  =  2E  sin  a. 

The  circulating  current  is 

Er  2E  sin  a        E  sin  a 


2Z 


where  Z  =  ZA  =  ZB  is  the  synchronous  impedance  of  each  of  the 
two  machines  which  are  assumed  to  be  similar.  Under  ordinary 
circumstances  the  current  I'  lags  nearly  90  degrees  behind  E'  ;  it 


.    .  _.        I     x, 

'l, 

"EB    (Referred  to  the  Load  Circuit) 
FIG.  333. 

is  approximately  in  phase  with  the  terminal  voltage  EA  of  A  and 
in  phase  opposition  to  EB]  and  it  therefore  transfers  power  from 
the  leading  to  the  lagging  machine  causing  the  leading  machine 
to  drop  back  in  phase  and  the  lagging  machine  to  come  up. 

The  component  of  I'  in  phase  with  Ef  represents  the  power 
consumed  by  the  passage  of  I'  through  the  resistance  of  the  local 
circuit  through  the  two  armatures  in  series;  the  component  in 
quadrature  behind  E'  represents  the  transfer  of  power  required 
to  keep  the  machines  in  step. 

Take  the  case  of  two  similar  60-cycle  alternators  driven  at 
240  r.p.m.  by  single-cylinder  steam  engines  of  1  per  cent,  speed 
variation.  The  driving  effort  and  the  speed  reach  a  maximum 
twice  during  the  revolution  and  fall  to  a  minimum  twice.  For 
one-quarter  of  the  revolution  the  alternator  A  runs  at  an  average 
speed  Y±  of  1  per  cent,  above  normal  and  it  therefore  moves 

9 


ahead  of  its  normal  position  by  T         =  Zn  °^  a  mechanical 


degree.  If  at  the  same  time  the  second  machine  B  is  running 
slow  it  will  fall  behind  its  normal  position  by  an  equal  angle  and 
the  maximum  phase  difference  between  the  two  machines  will  be 


SYNCHRONOUS  MACHINERY  349 

of  a  mechanical  degree.     These  alternators  would  have  30 
poles  and  therefore  the  maximum  phase  displacement  is 

20  X  \  =  20  X  T  =-6-75  electrical  degrees  =  2a. 


The  circulating  current  is  by  equation  (317) 

T'  =- 
=  Z 


£  sin  3.37°  =  I8C  sin  3.37° 


where  I8C  is  the  short-circuit  current  with  full  excitation  and  may 
be  from  two  to  six  times  full-load  current;  assuming  I8C  =  47  the 
circulating  current  is 

F  =  41  X  0.059  =  0.247 

and  is  approximately  25  per  cent,  of  full-load  current.  With 
lower  values  of  synchronous  reactance  and  consequently  higher 
values  of  7SC,  the  cross-currents  tend  to  become  very  large. 
Machines  with  very  small  synchronous  reactance  or  armature 
reaction  are  not  suitable  for  parallel  operation.  Too  high  syn- 
chronous reactance  reduces  the  cross-currents  and  the  synchroniz- 
ing power  too  much. 

These  power  cross-currents  when  of  great  magnitude  tend  to 
tear  the  machines  out  of  synchronism  and  they  also  cause  fluctua- 
tions of  the  voltage. 

If  the  two  machines  above  had  been  synchronized  so  that  their 
angular  velocities  rose  and  fell  together,  there  would  not  have 
been  any  power  currents  circulating  between  them  but  in  that 
case  the  frequency  of  the  voltage  of  the  system  would  pulsate 
and  power  currents  would  circulate  between  the  alternators  and 
any  synchronous  motors  operating  on  the  system. 

Alternators  driven  by  steam  engines  or  gas  engines  must  be 
provided  with  flywheels  large  enough  to  reduce  the  pulsations  of 
angular  velocity  during  the  revolution  to  a  negligible  amount. 

If  the  speed  characteristics  of  the  prime  movers  are  not  the 
same  and  the  speed  of  one  machine  tends  to  fall  below  the  other 
as  the  load  on  the  system  is  increased,  then  the  machine  driven 
by  the  prime  mover  of  closer  speed  regulation  takes  more  than 
its  share  of  the  load  and  so  relieves  the  other  machine  and  keeps 
its  speed  up. 

Thus  to  insure  a  proper  division  of  the  load  beween  alter- 
nators operating  in  parallel  it  is  necessary  that  their  prime  movers 


350  ELECTRICAL  ENGINEERING 

have  similar  speed  characteristics,  that  is,  that  their  speed  shall 
fall  under  load  by  the  same  amount  and  in  the  same  manner.  It 
is,  therefore,  preferable  that  the  prime  movers  have  drooping  speed 
characteristics.  The  voltage  characteristics  have  no  effect  on 
the  division  of  the  load  but  they  do  affect  the  amount  of  the 
reactive  cross-currents  between  the  machines. 

Let  R  and  X  represent  the  resistance  and  reactance  of  the 
circuit  through  two  alternators  in  parallel  including  the  lines  con- 
necting them  which  may  be  very  important  in  the  case  of  alter- 
nators in  different  stations. 

The  circulating  current  is 

E'          E'R-X  E'R          .    E'X 


R+jX        R*  +  X2 
the  quadrature  component  of  the  current  which  gives  the  power 


transfer  is  IQ  =  P2.i_  vV  "^'  ^^'  anc^  ^us  ^  ma^  ^e  seen 
reactance  or  armature  reaction  is  necessary  in  alternators  to  be 
operated  in  parallel. 

For  a  given  value  of  R  the  synchronizing  current  IQ  is  maxi- 
mum when  X  =  R]  this  result  is  obtained  by  differentiating  /0 
with  respect  to  X  and  equating  to  zero. 


dIQ      E'{(R2  +  Xz)  -  X(2X)} 

"  -        =  °'  for  maximum 


and  therefore  R2  -  X2  =  0,  or  X  =  R. 

Such  a  low  value  of  X  would  allow  excessive  currents  to  flow 
between  the  machines.  In  the  majority  of  cases  X  is  many  times 
greater  than  jR,  but  when  two  stations  are  to  be  operated  in 
parallel  R  and  X  may  approach  equality,  especially  with  under- 
ground distribution,  and  R  may  even  be  the  greater.  In  such  a 
case  the  operation  would  be  improved  by  inserting  extra  reactance 
to  make  X  =  R  and  so  increase  the  synchronizing  current. 

282.  Effect  of  Difference  of  Wave  Form.  —  If  two  machines 
in  parallel  are  adjusted  to  give  the  same  effective  value  of  voltage 
but  have  different  wave  shapes,  then,  since,  due  to  the  presence 
of  the  higher  harmonics,  the  voltages  are  not  equal  at  every 
instant,  reactive  cross-currents  will  flow  to  correct  these  in- 
equalities in  voltage.  These  currents  will  usually  be  very  small 
since  the  voltages  producing  them  are  small  and  they  are  of  high 
frequency  and  thus  the  path  through  the  two  machines  offers  a 


SYNCHRONOUS  MACHINERY        351 

high  impedance  to  them;  the  impedance  is,  however,  only  the  true 
impedance  and  not  the  synchronous  impedance. 

If  two  F-connected  alternators  with  neutrals  grounded  are 
connected  in  parallel  and  one  has  a  prominent  third  harmonic  in 
the  voltage  wave  of  each  phase  while  the  other  gives  a  true  sine 
wave,  then,  although  the  terminal  voltages  of  the  two  machines 
may  have  the  same  effective  values  and  the  same  wave  form,  still 
third  harmonic  circulating  currents  will  flow  between  them 
returning  through  the  neutral  connections. 

283.  Conclusions. — Three  kinds  of  circulating  currents  may 
exist  in  parallel  operation  of  alternators:  (1)  reactive  currents 
transferring  magnetization  between  the  machines  due  to  a  differ- 
ence in  excitation;  (2)  active  currents  transferring  power  between  - 
the  machines  due  to  phase  displacements  between  their  voltages ; 
and  (3)  higher-frequency  reactive  currents  due  to  differences  of 
wave  form. 

The  reactive  cross-currents  may  be  eliminated  by  regulating 
the  excitation,  that  is,  by  electrical  means,  but  the  power  cross- 
currents are  due  to  differences  in  speed  or  phase  and  can  be 
corrected  only  by  regulating  the  speed  of  the  prime  movers. 
Parallel  operation  of  alternators  is  therefore  not  so  much  an 
electrical  as  it  is  a  mechanical  problem. 

When,  therefore,  an  alternator  is  to  be  connected  in  parallel 
with  machines  supplying  the  load,  the  incoming  machine  must 
be  brought  up  to  synchronism  and  the  switch  connecting  it  to 
the  load  circuit  closed,  then  the  governor  of  its  prime  mover 
must  be  adjusted  so  that  it  supplies  its  proper  share  of  the  load 
and  its  field  must  be  adjusted  so  that  it  supplies  its  proper  share 
of  the  reactive  current  required  by  the  load.  Ordinarily  the 
machines  will  be  operated  at  approximately  the  same  power  fac- 
tor. Further  adjustment  may  be  necessary  as  the  load  changes. 

284.  Hunting. — If  two  alternators  are  operating  in  parallel 
and  one  drops  behind  the  other  in  phase  due  to  a  sudden  decrease 
in  the  speed  of  its  prime  mover,  the  second  machine  supplies 
power  to  pull  it  into  synchronism  again.     The  impulse  received 
causes  it  to  swing  past  its  mean  position  and  it  oscillates  a  few 
times  before  falling  into  step. 

If  the  action  producing  the  speed  pulsation  is  repeated  periodic- 
ally and  coincides  with  the  natural  period  of  the  machine  the 
oscillations  instead  of  dying  out  will  increase  in  amplitude  until 
they  are  limited  by  the  losses  in  the  pole  faces  and  the  dampers  or 


352  ELECTRICAL  ENGINEERING 

until  the  machines  fall  out  of  step.  When  the  oscillations  tend  to 
become  cumulative  the  machines  are  said  to  be  hunting. 

Hunting  may  occur  in  a  similar  way  in  the  case  of  a  synchro- 
nous motor  supplied  by  an  alternator.  If  the  load  on  the  motor 
suddenly  increases  it  falls  back  in  phase  to  receive  the  extra 
power  required  and  oscillates  about  its  final  phase  position  before 
running  in  synchronism  again.  This  oscillation  may  become  con- 
tinuous as  in  the  case  of  alternators  in  parallel. 

The  amplitude  of  the  oscillations  in  hunting  is  very  much  re- 
duced by  the  use  of  dampers  in  the  form  of  grids  or  squirrel-cage 
windings  placed  in  slots  in  the  pole  faces  on  the  rotor.  At 
synchronous  speed  the  armature  reaction  flux  is  stationary  rela- 
tive to  the  fields  and,  therefore,  does  not  produce  any  current  in 
the  grids  but  if  the  machine  falls  below  or  runs  above  synchro- 
nous speed,  the  flux  sweeps  across  the  grids  and  produces  e.m.fs.  in 
them  and  large  currents  flow  which  react  on  the  field  and  tend  to 
hold  the  machine  exactly  in  synchronism. 

Dampers  are  applied  to  alternators  only  in  cases  where  hunting 
is  liable  to  occur,  due  to  the  fact  that  the  prime  movers  are  re- 
ciprocating engines  or  gas  engines  but  they  should  be  used  on  all 
large  synchronous  motors  since  these  are  usually  high-speed 
machines  with  small  moment  of  inertia  and  a  high  natural  fre- 
quency of  oscillation  which  is  more  liable  to  coincide  with  some 
forced  frequency  on  the  system.  The  dampers  are  also  required 
to  make  the  motors  self-starting. 

285.  Frequency  of  Hunting. — Referring  to  Fig.  333,  the  power 
supplied  by  A  to  pull  B  up  in  phase  again  is 

P8  =  El'  cos  «, 
but 

7/  _  E'     _  2E  sin  a  _  E    . 

~2Z8"  ~^zT~  ~^S]   a' 

and  thus  the  synchronizing  power  is 

Ez   .  E2  'E2   , 

Pa  =  •=-  sin  a  cos  a  =  p-^-  sin  2a  =  ^r  2a, 
/js  £*i*  ALS 

since  for  small  angles  the  sine  is  approximately  equal  to  the  angle, 
here  a  is  expressed  in  electrical  radians. 
Changing  a  into  mechanical  radians, 

(318) 


SYNCHRONOUS  MACHINERY  353 

The  synchronizing  torque  is 


P8 33,000  =         2ZS      _      33,000  = 

1  s  ~  27T  X  r.p.m.         746          2^  X  r.p.m.         746 

3.52  E*p 

Z8  X  r.p.m.  ° 
and  it  is  directly  proportional  to  the  angular  displacement  a. 

Since  the  sum  of  the  moments  of  the  external  forces  acting  on 
a  rigid  body  is  equal  to  the  moment  of  inertia  of  the  body  about 
the  axis  of  rotation  multiplied  by  its  angular  acceleration,  the 
equation  for  the  resulting  motion  may  be  expressed  as 

T        j^a  3.52  E2p  ,qQv 

—    1  8    =    1   375-    =    —  'rjr— Oi.  (3lV) 

at*  Ls  X  r.p.m. 

where  /  is  the  moment  of  inertia  of  the  rotating  field  member 
including  the  flywheel.     The  negative  sine  is  used  because  the 
torque  tends  to  decrease  the  angular  displacement  a. 
This  equation  may  be  written 

/J2  Q    KO    77'2P 

II   CX.  O.iJA  -C/  i  7  „  /or»rk\ 

W  =  "  Z I  X  r     m  a  =     ~         '  ^       ^ 

This  is  the  equation  of  a  simple  harmonic  motion  of  periodic 
time 

—  sec.  (321) 


and  therefore  the  natural  frequency  of  hunting  is 


,         1         1 
fk  =        = 


The  frequency  of  hunting  is  directly  proportional  to  the  voltage 
E  or  to  the  air-gap  flux  3>  and  is  inversely  proportional  to  the 
square  root  of  the  moment  of  inertia  /  and  the  synchronous 
impedance  of  the  armature  and  connecting  feeders. 

The  natural  frequency  may  be  increased  by  increasing  E, 
that  is,  increasing  the  voltage  of  the  system  but  this  is  not  usu- 
ally practicable  and  it  may  be  decreased  by  inserting  reactance 
in  the  lines  between  the  machines. 

The  natural  frequency  of  oscillation  of  an  alternator  should 
not  approach  within  20  per  cent,  of  any  forced  frequency  which  is 
liable  to  occur  on  the  system  and  the  flywheel  is  usually  designed 
so  that  it  increases  /  to  such  a  value  that  the  natural  frequency 
is  at  least  20  per  cent,  below  the  lowest  forced  frequency. 

The  same  formula  holds  in  the  case  of  a  synchronous  motor 

23 


354  ELECTRICAL  ENGINEERING 

operating  on  a  large  constant-voltage  system.  E  may  be  taken  as 
the  motor  voltage  which  is  dependent  on  the  excitation.  When 
the  motor  is  under-excited  E  is  low  and  the  synchronizing  power 
transfer  per  degree  angular  displacement  is  small  and  the  natural 
period  of  oscillation  is  low.  The  electromagnetic  coupling  is  soft 
and  the  motor  may  be  thrown  out  of  step  by  a  sudden  increase 
of  load.  When  the  motor  is  over-excited  the  synchronizing 
power  is  large  and  the  natural  frequency  is  high.  The  electro- 
magnetic coupling  is  stiff  and  the  motor  is  stable  in  operation. 
If,  however,  any  forced  frequency  occurs  approximating  the 
natural  frequency,  serious  hunting  will  occur. 

Hunting  may  sometimes  be  reduced  or  eliminated  by  changing 
the  field  excitation  and  thus  changing  the  natural  period  of  oscil- 
lation of  the  machine. 

Machines  with  high  armature  reaction  are  much  less  liable  to 
hunt  than  machines  of  low  armature  reaction  since  the  high  arm- 
ature reaction  reduces  the  circulating  currents  produced  by 
changes  in  phase  and  lowers  the  natural  frequency  of  oscillation. 

286.  Design  of  Alternating-current  Generators  and  Motors. — 
The  design  of  alternating-current  generators  and  synchronous 
motors  is  similar  to  that  of  direct-current  generators  and  motors 
but  without  the  difficulties  and  limitations  due  to  commutation. 
Much   larger  outputs,   higher  voltages   and   greater  peripheral 
speeds  must  be  provided  for  in  certain  cases.     The  voltage  regu- 
lation of  alternators  is  not  so  good  as  that  of  direct-current  gen- 
erators, since  a  large  component  of  the  generated  voltage  is  con- 
sumed by  the  armature  reactance.     The  regulation  also  depends 
on  the  power  factor  of  the  load.     To  obtain  reasonable  values  of 
regulation  the  air  gap  must  be  made  long  and  therefore  the  ratio 
of  field  ampere-turns  to  armature  ampere-turns  must  be  large. 
For  machines  of  very  large  output  as  turbo-alternators  the  periph- 
eral speed  becomes  a  limiting  factor,  the  frame  must  be  made 
long  and  forced  ventilation  is  necessary  to  prevent  undue  rise  of 
temperature. 

Following  are  some  of  the  formulae  and  constants  involved. 

287.  Electromotive  Force  Equation. 

E.m.f.  per  phase  =  E  =  4dyfn<j>W-*      (Art.  248) 

or  assuming  a  distribution  factor  6  =  0.96  and  a  form  factor 
7  =  1.11,  the  e.m.f.  is 

E  =  4.26/n4>10-8  volts.  (323) 


S  YNCHRONO  US  MA  CHINER  Y 


355 


288.  Output  Equation. 

Output  in  volt-amperes  =  p'EI  (where  p'  =  number  of  phases) 


=  4.26p'f<t>nl  10~8  =  4.26    X 


r.p.m.  X  p 
120 

X  Bt 


X 


X  10-8 


4.26 


120  X  2 

4.26 
120  X  2 


or 


Ba  = 


X  10-8  X  TT  X  r.p.m.  X  Ba^q  X  prX  Dale 
X  10~8  X  7T2  X   r.p.m.  X  Bg^q  X  Da2Lc, 

_  volt-amperes  X  5.7  X  108. 
r.p.m.  X  Bg^q 

average   gap   density,    varies   from   40,000   to 


(324) 


60,000  lines  per  square  inch,  but  is  usually  below  50,000. 
^  =  pole  enclosure  =  0.60  to  0.65. 
q  =  ampere  conductors  per  inch,  ranges  from  400  to  1,000;  a 

good  average  value  is  500. 

Lc  .    frame  length 

—  =  ratio j ;    ,      ranges  from  0.6  to  2.0;  a  good  average 

is  1. 

p  =  number  of  poles  =  is  fixed  by  the  frequency  and 

the  r.p.m. 

Approximate  rated  speed  in  r.p.m.  of  prime  movers  for  gener- 
ators of  250-,  500-  and  1,000-kw.  capacity. 


Prime  mover 

Rated  speed  in  r.p.m. 

250-kw. 

500-kw. 

1,000-kw. 

Moderate-speed  steam  engine  
High-speed  gas  engine  

160 
250 
350 
1,000 
2,500 
3,000 

120 

200 
270 

1,800- 
2,500 

80 
120 
220 

1,200 
1,900 

High-speed  steam  engine 

De  Laval  steam  turbine  

Curtis  steam  turbine  

Parsons  steam  turbine 

The  speed  of  hydraulic  turbines  is  controlled  by  the  head  of 
water  available. 

Peripheral  speed  varies  over  a  wide  range;  for  low-speed  ma- 
chines a  good  average  value  is  3,500  ft.  per  minute  but  in  very 
large  turbo-alternators  it  may  be  as  high  as  25,000  ft.  per  minute- 


356 


ELECTRICAL  ENGINEERING 


289.  Flux  Densities. — The  following  table  from  the  "Standard 
Handbook"  gives  the  average  flux  densities  used  in  the  various 
parts  of  the  magnetic  circuit. 


Part 

Frequency 

Flux  density  in  lin 

es  per  square  inch 

Ordinary  iron 

Silicon  steel 

Armature  core  .... 

. 

25  cycles  per  second;  60,000  to  70,000 

70,000  to  80,000 

60  cycles  per  second 

50,000  to  60,000 

60,000  to  70,000 

Armature  teeth...  . 

25  cycles  per  second 

100,000  to  120,000 

60  cycles  per  second 

90,  000  to  110,000 

Pole  core  

95,000  to  110,000  with  wrought  iron  or  steel  poles. 

VnVp 

90,000  to  100,000  for  cast  steel. 

30,000  to  35,000    for  cast  iron. 

290.  Current  Densities. — Current  densities  in  the  armature 
conductors  range  from  1,500  to  2,500  amp.  per  square  inch;  the 
lower  values  should  be  used  in  machines  which  are  difficult  to 
cool,  that  is,  machines  with  long  cores  or  in  high-voltage  machines 
in  which  the  thick  slot  insulation  restricts  the  radiation  of  the 
heat  due  to  the  copper  loss.  Short  machines  of  large  diameter 

are  easy  to  cool  since  the  end  connec- 
tions are  spread  out  and  have  a  good 
radiating  surface. 

Current  density  in  field  windings 
varies  from  1,200  to  2,500  amp.  per 
'square  inch  depending  on  the  con- 
struction, insulation,  available  space 
and  on  the  temperature  rise  allowed. 

291.  Insulation  for  High-voltage 
Alternators. — The  very  best  materials 

must  be  used  because  the  space  is  limited.  Micanite  or  Bakel- 
ized  paper  tubes  may  be  used  or  the  insulation  may  be  built 
up  of  a  number  of  layers  of  empire  cloth  or  similar  material. 
The  curve  in  Fig.  334  shows  approximately  the  thickness  of  the 
slot  insulation  for  various  voltages.  The  built-up  insulations 
may  require  more  space  than  indicated  by  the  curve, 


0.3 

^^ 

o 
fl 

«(!•> 

^^ 

5°*2 
5 

/ 

X 

Thickii 
j 

/ 

0  5000         10000        15000       20000 

Effective  E.M.F.  in  Volts 

FIG.  334. — Thickness  of  slot 

insulation  for  alternators. 


SYNCHRONOUS  MACHINERY        357 

The  winding  should  not  be  distributed  in  more  slots  than 
is  necessary  since  so  much  of  the  slot  width  is  required  for 
insulation. 

292.  Extra  Insulation  Required  under  Special  Conditions.  — 
If  a  three-phase,  star-connected,  11,000-volt  generator  is  operated 
with  its  neutral  grounded  through  a  resistance  of  10  ohms,  and  a 
ground  occurs  on  one  line,  the  short-circuit  current  flowing  through 
the  ground  resistance  will  raise  the  potential  of  the  neutral  point 
and  will  seriously  increase  the  potential  stresses  from  the  other 

11  000 
windings  to  the  core.     The  voltage  per  phase  is  —  ~=-  =  6,350 

v  3 

volts  and  taking  the  resistance  in  the  short-circuit  as  12  ohms  the 

6  Q  FAQ 
current  is     '         =  530  amp.     This  will  raise  the  neutral  point 

to  530  X  10  =  5,300  volts  above  ground  and  may  increase  the 
stress  from  the  ungrounded  phases  to  the  core  to  approximately 
6,350  +  5,300  -  11,650  volts. 

293.  Armature  Windings.  —  Armature  windings  are  discussed 
in  Arts.  241  to  245.     They  should  be  distributed  rather  than 
concentrated  to  dissipate  the  heat  more  readily  and  to  improve 
the  wave  form  and  reduce  the  reactance.     Full-pitch  windings 
give  the  highest  voltage  but  fractional-pitch  windings  as  low  as 
66  or  even  50  per  cent,  of  full  pitch  are  used  to  eliminate 
objectionable  harmonics  or  to  reduce  the  length  of  the  end 
connections. 

294.  Slots  per  Pole.  —  Large  slow-speed  machines  with  many 
poles  have  one  to  three  slots  per  phase  per  pole,  while  high-speed 
machines  have  from  four  to  nine  slots  per  phase  per  pole. 

_.     .     tooth  width 

Ratio      ,   ,  —  .  ,,,      vanes  from  1  to  2  for  open  slots  and  the 
slot  width 

slot  depth 

ratl°  vanes  from  2  to  3-5  or  4- 


295.  Regulation.  —  The  voltage  regulation  depends  on:  (1) 
armature  resistance,  (2)  armature  reactance,  (3)  armature  reac- 
tion and  (4)  load  power  factor. 

1.  Armature  resistance  is  determined  more  from  considerations 
of  current  density  and  temperature  rise  than  of  regulation. 

2.  The  armature  reactance  drop  is  produced  by  the  various 
leakage  fluxes  and  depends  on  the  shape  of  the  slots,  the  length  of 
the  gap,  the  pitch  and  distribution  of  the  winding  (see  Arts.  252 
and  253). 


358  ELECTRICAL  ENGINEERING 

3.  The   effect   of   armature   reaction   is   determined   by   the 

field  ampere-turns 

>  armature  ampere-turns'  the  length  of  the  ^  and  the 
flux  densities  in  the  gap,  the  teeth  and  the  poles.     The  ratio- 
field  ampere-turns 

-  usually   lies   between   2.0   and   3.0,    the 
armature  ampere-turns 

lower  value  being  for  high-speed  machines. 

4.  The  effect  of  load  power  factor  or  regulation  is  discussed  in 
Art.  261. 

For  the  best  regulation  use  a  high  air-gap  density,  a  long  gap 

field  ampere-turns 

and  a  large  ratio  -  —  • 

armature  ampere-turns 

Air  gaps  are  much  longer  than  in  direct-current  machines 
ranging  from  0.5  or  lower  to  2.5  in.  in  some  turbo-alternators. 
Eighty-five  or  90  per  cent,  of  the  field  m.m.f.  may  be  required 
for  the  gap. 

Short  gaps  make  the  regulation  poor  and  result  in  distortion 
of  the  gap  flux  and  the  wave  of  generated  e.m.f. 

Ordinary  values  of  regulation  are  5  to  8  per  cent,  at  100  per  cent, 
power  factor  and  15  to  25  per  cent,  at  80  per  cent,  power  factor. 

With  automatic  voltage  regulators,  close  regulation  is  not 
necessary  and  may  be  a  distinct  disadvantage  since  the  armature 
current  under  conditions  of  short-circuit  is  likely  to  reach  dan- 
gerous dimensions  in  large  low-reactance  alternators. 

296.  Excitation  Regulation. — Excitation  regulation  or  the  per 
cent,  increase  of  field  excitation  to  maintain  constant  voltage 
from  no  load  to  full  load  ranges  from  10  to  15  per  cent,  for  non- 
inductive  load  to  25  to  35  per  cent,  for  80  per  cent,  power-factor 
load. 

297.  Excitation. — Alternators    and    synchronous    motors    are 
excited  by  a  separate  direct-current  shunt  generator  called  an 
exciter,  usually  at  125  or  250  volts. 

The  exciter  should  preferably  be  driven  by  a  separate  prime 
mover  to  reduce  the  variation  of  the  alternator  voltage  with 
speed.  When  the  exciter  is  mounted  on  the  same  shaft  as  the 
alternator  a  1  per  cent,  decrease  in  speed  produces  more  than  1 
per  cent,  decrease  in  the  exciter  voltage  and  more  than  2  per  cent, 
decrease  in  the  alternator  voltage. 

The  rating  of  the  exciter  is  usually  from  1  to  2  per  cent,  of  the 
alternator  rating  unless  it  is  necessary  to  maintain  the  voltage 
to  very  low  power  factors,  when  it  must  be  larger. 


SYNCHRONOUS  MACHINERY  359 

298.  Losses. — The  losses  in  alternators  are  similar  to  those  in 
direct-current  generators. 

The  field  copper  loss  is  usually  greater  than  in  direct-current 
machines  since  for  good  regulation  the  field  must  be  stronger. 
This  loss  is  about  1  per  cent,  for  non-inductive  loads  but  will  be 
greater  for  inductive  loads. 

Armature  copper  loss  is  similar  to  that  in  direct-current  ma- 
chines but  for  the  large  outputs  conductors  of  large  section  must 
be  used  and  they  must  be  laminated  to  prevent  large  extra  losses 
due  to  eddy  currents  as  discussed  in  Art.  188.  Eddy-current 
and  hysteresis  losses  in  the  teeth  and  other  metal  parts  near  the 
conductors  which  are  due  to  the  presence  of  the  alternating  cur- 
rent in  the  conductors  add  to  the  copper  loss  and  increase  the 
apparent  or  effective  resistance  of  the  armature.  In  some  cases 
it  may  be  50  per  cent,  greater  than  the  true  ohmic  resistance. 

The  no-load  core  loss  may  be  from  one  and  a  half  to  four  times 
the  full-load  copper  loss  depending  on  the  rated  speed  and  fre- 
quency. With  silicon -steel  this  loss  may  be  reduced  by  30  or  40 
per  cent.  The  increase  of  core  loss  under  load  may  be  as  great 
as  25  or  50  per  cent,  in  some  cases. 

Windage  and  bearing-friction  losses  in  moderate-speed  ma- 
chines range  from  0.5  to  1  per  cent,  but  in  high-speed  turbo-alter- 
nators with  fans  for  cooling  this  loss  may  reach  1.5  per  cent. 

299.  Ventilation. — The  rate  at  which  heat  is  dissipated  from 
a  surface  is  proportional  to  the  difference  in  temperature  between 
the  surface  and  the  cooling  air.     A  continuous  circulation  of  cool 
air  is  required  to  prevent  dangerous  temperature  increases  in  gen- 
erators and  motors. 

There  is  very  little  difficulty  experienced  in  cooling  slow-speed 
machines  as  the  losses  per  pound  of  material  are  small  and  the 
radiating  surfaces  are  large.  The  rotor  with  its  salient  poles  acts 
as  a  fan  and  blows  air  over  the  armature  core  and  end  connec- 
tions. If  necessary,  fans  may  be  added  to  direct  the  flow  of  air 
and  increase  the  cooling  effect. 

The  insulation  on  the  coils  is  a  poor  heat  conductor  and  most 
of  the  heat  generated  by  the  copper  losses  passes  along  the  con- 
ductors to  the  end  connections  where  the  insulation  is  thinner 
and  is  there  radiated  to  the  air.  The  end  connections  must  be 
very  well  ventilated. 

When  the  frame  length  is  small  and  the  diameter  is  large,  the 
end  connections  are  spread  out  and  the  machine  is  easy  to  cool. 


360 


ELECTRICAL  ENGINEERING 


When  the  frame  length  is  great  and  the  diameter  is  small  there 
is  very  little  space  for  the  end  connections  and  they  are  difficult 
to  cool.  Radial  vent  spaces  are  left  in  the  core  by  placing  special 
vent  plates  between  blocks  of  punchings  at  suitable  intervals. 
The  vent  spaces  must  be  wide  enough  to  allow  the  air  to  pass 
through  them  freely.  The  cooling  air  carries  off  the  heat  from 
the  iron  due  to  the  core  losses  and  also  part  of  that  due  to  the 
copper  losses.  The  remainder  of  the  heat  due  to  the  core  losses 
is  carried  along  the  laminations  to  the  air  gap  or  to  the  frame. 
This  is  called  radial  ventilation  (Figs.  242  and  336). 

Another  method  of  cooling  makes  use  of  axial  ducts  instead  of 
or  supplementing  the  radial  ducts  (Fig.  243).  The  axial  ducts 
supply  air  which  passes  over  the  edges  of  the  punchings  and  is 
very  effective  as  a  cooling  agent,  since  the  iron  core  conducts  heat 
from  20  to  100  times  better  along  the  laminations  than  across 
them,  depending  on  the  method  of  stacking. 

High-speed,  steam  turbine-driven  alternators  of  large  out- 
put are  very  difficult  to  cool.  To  limit  the  peripheral  speed 


Steel  Line. 


Two  Pole,  Cylindrical  Eotor 


(5) 

Four  Pole,  Cylindrical  Eotor 

FIG.  335. 


Two  Pole,  Cylindrical  Rotor 


the  diameter  must  be  made  small  and  consequently  the  frame 
must  be  long.  Forced  ventilation  is  necessary  and  the  machines 
must  be  completely  enclosed  to  give  control  of  the  cooling  air 
and  to  prevent  objectionable  noise.  Openings  are  provided  for 
the  passage  of  cooling  air.  Fans  are  placed  on  the  rotor  and 
the  air,  after  it  has  been  cleaned  and  cooled,  is  drawn  along  the 
air  gap  and  forced  out  through  the  stator  radial  ducts  into  pas- 
sages which  carry  it  to  the  outlet.  Axial  ducts  may  be  provided 
in  the  core  if  it  is  considered  necessary. 

The  rotor  with  its  small  diameter  and  restricted  space  for  field 
copper  is  difficult  to  cool.     Most  of  the  heat  developed  in  it  is 


SYNCHRONOUS  MACHINERY 


361 


carried  to  the  air  gap  and  dissipated  there.  In  some  cases  axial 
ducts  are  cut  at  the  bottom  of  the  rotor  slots  and  these  are  very 
effective  (Fig.  335). 

About  100  cu.  ft.  of  cooling  air  per  minute  are  required  per 
kilowatt  lost  to  limit  the  temperature  rise  at  full  load  to  permis- 
sible values. 

For  example,  in  a  25,000-kva.  turbo-alternator  of  97  per 
cent,  efficiency  the  losses  amount  to  750  kw.  and  about  75,000 
cu.  ft.  of  air  per  minute  are  required.  With  a  velocity  of  5,000  ft. 

75  000 
per  minute  the  section  of  the  air  passages  must  be    ,  '     -  =  15 

O,UUU 

sq.  ft.  Air  may  be  admitted  from  both  ends  but  even  so  the 
gap  must  be  made  very  long  to  give  the  required  area.  The  gap 
length  in  some  cases  must  be  determined  in  this  way. 

In.  the  circumferential  system  of  ventilation  air  is  admitted  at 
one  side  of  the  machine  and  passes  circumferentially  through  the 
stator  ducts  to  the  other  side  where  it  is  discharged.  Air  must 
at  the  same  time  be  forced  along  the  air  gap  to  cool  the  rotor 
(Fig.  336). 


Stator  Ducts 


FIG.  336. — Circumferential  ventilation. 


300.  Cylindrical  Rotors. — The  rotors  of  steam-turbine  alter- 
nators present  new  features  in  design.  To  keep  the  peripheral 
speeds  down  to  safe  values,  that  is,  below  25,000  ft.  per  minute, 
the  diameter  must  be  made  small  and  the  length  great.  The 
field  winding  is  distributed  and  must  be  completely  imbedded  in 
slots.  The  rotor  teeth  must  be  designed  with  sufficient  strength 
to  stand  the  enormous  centrifugal  forces  acting  on  them.  At 
25,000  ft.  per  minute  a  single  pound  of  material  is  acted  upon  by 
a  force  of  over  1  ton.  Fig.  335  shows  three  sections  of  cylindrical 


362  ELECTRICAL  ENGINEERING 

rotors.  The  four-pole  rotor  in  (6)  is  ventilated  by  means  of  axial 
ducts  below  the  slots.  A  section  of  the  winding  in  one  slot  is 
shown.  To  force  the  winding  solidly  into  place  pressure  is  ap- 
plied to  the  bronze  wedge  and  then  the  steel  liners  are  driven  into 
place. 

Since  the  space  for  the  field  winding  is  limited  high  cur- 
rent densities  must  be  used  and  the  rotors  run  hot.  Only  those 
materials  which  can  stand  high  temperatures  must  be  used  for 
insulation. 


CHAPTER  X 


TRANSFORMERS 

301.  The  Constant-potential  Transformer. — The  constant- 
potential  transformer  consists  of  one  magnetic  circuit  interlinked 
with  two  electric  circuits,  the  primary  circuit  which  receives 
energy  and  the  secondary  circuit  which  delivers  energy.  Its  func- 
tion is  to  transform  electric  power  from  low  voltage  and  large 
current  to  high  voltage  and  small  current,  or  the  reverse.  In 
step-up  transformers  the  primary  is  the  low-voltage  (L.V.)  side 
and  the  secondary  is  the  high-voltage  (H.V.)  side.  In  step- 
down  transformers  the  primary  is  the  high-voltage  side. 

In  the  following  discussion  letters  with  the  subscript  1  will  be 
used  to  represent  primary  quantities  and  with  the  subscript  2  to 
represent  secondary  quantities. 


Primary 

•         *° 

^           Core 

s? 

—  •**< 
««— 

«•••—• 

Secondary 
I 

>       E2 
l 

—  — 
—  — 

,  < 

'  < 

,  < 

Ei       E'< 

jg 

/ 

FIG.  337. — Constant   potential    transformer. 


FIG.  338. 


Fig.  337  represents  a  transformer.  The  core  is  made  up  of 
thin  sheets  of  iron  or  steel  of  high  permeability  with  small  hystere- 
sis and  eddy-current  loss. 

The  primary  winding  consists  of  n\  turns  in  series  and  has  a 
resistance  of  r\  ohms,  a  self -inductive  or  leakage  reactance  of  x\ 
ohms  and  thus  an  impedance  of  Zi  =  \A"i2  +  #i2  ohms. 

The  secondary  winding  consists  of  n2  turns  in  series.  Its  re- 
sistance is  r2  ohms,  its  reactance  is  x2  ohms  and  its  impedance  is 
Z2  =  Vr22  +  ^22  ohms 

363 


364  ELECTRICAL  ENGINEERING 

When  an  alternating  e.m.f.  EI  is  impressed  on  the  primary 
winding  with  the  secondary  open,  a  current  /o  flows  in  the  pri- 
mary and  produces  an  alternating  flux  through  the  core  of  maxi- 
mum value  <£.  The  current  Jo  is  called  the  exciting  current  of 
the  transformer  and  consists  of  two  components  (Fig.  338)  IM 
in  phase  with  the  flux  <£,  called  the  magnetizing  current,  and  Ic 
in  quadrature  ahead  of  the  flux  and  in  phase  with  the  impressed 
e.m.f.,  called  the  core-loss  current.  The  product  of  EI  and  Ic 
is  the  power  wasted  in  the  core  loss,  that  is,  in  supplying  the 
hysteresis  and  eddy  current  losses  of  the  transformer.  The  ex- 
citing current,  therefore,  lags  by  an  angle  00,  which  is  less  than 
90  degrees,  behind  the  impressed  e.m.f.  Cos  00  is  the  no-load 
power  factor  of  the  transformer. 

The  excitng  current  /o  is  from  3  to  10  per  cent,  of  full-load 
current  and  the  no-load  power  factor  is  of  the  order  of  30  per 
cent. 

The  alternating  flux  produced  by  the  magnetizing  current 
links  with  the  secondary  winding  and  induces  in  it  an  e.m.f., 

dd> 
e2  =  —  nz  -7r  10~8  volts. 

If  the  frequency  is  /  cycles  per  second  and  the  flux  follows  a 
sine  wave  of  maximum  value  <l>,  the  instantaneous  e.m.f.  induced 
in  the  secondary  is 

e2  =  —  n2  -j  ($  sin  2-7r/010~8  volts 

=  -  2irfnz3>  cos  27r/<10"8  volts 

2irfn23>  sin  (2irft  -  90)  10~8  (325) 

this  is  a  sine  wave  of  e.m.f.  in  quadrature  behind  the  flux,  of 
maximum  value 

#2max.    =    2irfn23>  10~8 

and  effective  value 


10~8  =  4.44/n2$  10~8  volts.  (326) 

v  2 

The  flux  also  links  with  the  primary  winding  and  induces  in  it 
an  e.m.f.  of  instantaneous  value 


10-8  sin  (2irft  -  90), 
a  sine  wave  of  maximum  value 

10~8 


TRANSFORMERS  365 

and  effective  value 


Eb  =  4.44/n^  10-8  volts.  (327) 

This  e.m.f.  induced  in  the  primary  is  almost  equal  in  value  and 
opposite  in  phase  to  the  impressed  e.m.f.,  the  vector  sum  of  the 
two  being  the  small  component  of  impressed  e.m.f.  required  to 
drive  the  exciting  current  through  the  impedance  of  the  primary 
winding.  Thus 


This  component  has  been  neglected  in  Fig.  338.  The  induced 
e.m.fs.  Eb  and  E%  are  directly  in  phase  since  they  are  produced 
by  the  same  flux,  and  their  intensities  are  in  the  ratio  of  the  turns 
on  the  two  windings;  therefore, 

lO-8       nx          ....  ,-OQ. 

m  «  =  ~~  =  ratl°  of  turns.  (328) 

10~8       n2 

If  the  secondary  is  connected  to  a  receiver  circuit  of  impedance 
Z  =  R  +  jXy  a  current  I  \  flows  in  it.  The  primary  current  is  at 
the  same  time  increased  by  a  component  /',  the  primary  load 
current,  -which  exerts  a  m.m.f.  equal  and  opposite  to  that  of  the 
secondary  current. 
Thus 

nil'  =  n2/2- 
and 

Y,  =  —  =  ratio  of  transformation  (329) 

J.  11/2 

The  resultant  m.m.f.  acting  on  the  magnetic  circuit  of  the  trans- 
former is  still  that  of  the  primary  exciting  current  and  the  flux 
threading  the  two  windings  remains  almost  constant. 

The  primary  current  under  load  is  I\  and  has  two  components 
/o  the  exciting  current,  which  is  proportional  to  the  flux,  and  /' 
the  load  current  which  is  proportional  to  the  secondary  current. 

The  exciting  current  /o  can  be  expressed  as  the  product  of  the 
primary  induced  e.m.f.  Eb  and  the  primary  exciting  admittance 
YQ  =  go  -  jb0;  thus 

7o  =  Eb  (go  -  jbo)  =  E'  (go  -  jb0),  (330) 

where  Ef  is  the  component  of  impressed  e.m.f.  required  to  over- 
come the  induced  e.m.f.  Eb. 

The  primary  load  current  is  /'  =  —  1  2,  and  is  opposite  in  phase 
to/o. 


366  ELECTRICAL  ENGINEERING 

As  the  load  on  the  transformer  is  increased,  the  primary  in- 
duced e.m.f.  decreases  (except  when  the  power  factor  of  the  load 
is  leading)  because  a  larger  component  of  impressed  e.m.f.  is 
consumed  in  driving  the  current  through  the  primary  impedance, 
thus, 

E'  =  -  Eb  =  E,  -  IiZi.  (331) 

A  smaller  flux,  is,  therefore,  required  and  a  smaller  exciting 
current.  The  decrease  in  flux  from  no-load  to  full-load  non-in- 
ductive is  1  or  2  per  cent,  and  for  an  inductive  load  of  50  per  cent, 
power  factor  is  only  5  or  6  per  cent.  With  anti-inductive  load  the 
flux  increases. 

The  secondary  induced  e.m.f.,  which  is  proportional  to  the 
primary,  decreases  with  it. 

The  secondary  terminal  e.m.f.  E  is  less  than  the  secondary 
induced  e.m.f.  by  the  e.m.f.  consumed  by  the  secondary  imped- 
ance thus 

E  =  E2  -  I2Z2.  (332) 


FIG.  339.  —  Vector  diagram  of  a  transformer  with  inductive  load. 

302.  Vector  Diagrams  for  the  Transformer.—  Fig.  339  is  the 

vector  diagram  of  e.m.fs.  and  currents  in  a  transformer  with  a 
load  impedance  Z  =  R  +  jX  and  a  load  power  factor  cos  6  = 


E  =  secondary  terminal  e.m.f. 

/2  =  secondary  current  lagging  behind  E  by  angle  0. 
2  =  e.m.f.  consumed  by  secondary  resistance,  in  phase  with 

/2. 

2^2  =  e.m.f.  consumed  by  secondary  reactance,  in  quadra- 

ture ahead  of  72. 
-^2  =  e.m.f.  consumed  by  secondary  impedance. 


TRANSFORMERS 


367 


EI  =  e.m.f.  induced  in  the  secondary  winding  by  the  alter- 
nating flux  <£.      #2  =  jp  +  /20"2  +  j%z). 

Eb  =  e.m.f.  induced  in  the  primary  winding  by  the  alter- 

r>.-, 

nating  flux  $ 


- 
b  - 


$  =  flux  linking  both  primary  and  secondary  windings,  in 

quadrature  ahead  of  Eb  and  EZ. 
Io  =  primary  exciting  current  leading  the  flux  <£  by  an  angle 

a  =  90  —  00,  where  cos  do  is  the  primary  power  factor 

at  no  load.~~ 
/'  =  primary  load  current  in  phase  opposition  to  72,  /'  = 


I\  =  total  primary  current. 


FIG.  340. — Transformer    with    non-  FIG.  341. — Transformer  with    a     load 

inductive  load.  power  factor  of  50  per  cent,  leading. 


Ef  = 


=  e 


1  = 


62  = 


component  of  primary  impressed  e.m.f.  required  to 
overcome  the  primary  induced  e.m.f  Eb. 
s.m.f.  consumed  by  primary  resistance,  in  phase  with 

r 

e.m.f.  consumed  by  primary  reactance,  in  quadrature 
ahead  of  /i. 

e.m.f.  consumed  by  primary  impedance, 
primary  impressed  e.m.f.     EI  =  E'  +  li(r\  +  jxi). 
angle  of  lag  of  the  primary  current  behind  the  primary 
impressed  e.m.f. ;  cos  61  =  primary  power  factor, 
angle  of  lag  of  the  secondary  current  behind  the  sec- 
ondary induced  e.m.f. 


368 


ELECTRICAL  ENGINEERING 


Fig.  340  shows  the  vector  diagram  of  the  transformer  with  a 
non-inductive  load  and  Fig.  341  with  a  capacity  load  of  50  per 
cent,  power  factor  leading. 

The  secondary  current  and  secondary  terminal  e.m.f.  are  the 
same  in  the  three  cases  and  it  may  be  seen  that  the  required  pri- 
mary impressed  e.m.f.  is  greatest  in  the  case  of  the  inductive 
load  and  least  in  the  case  of  the  capacity  load. 

303.  Exciting  Current. — When  a  sine  wave  of  e.m.f.  is  im- 
pressed on  the  primary  winding  of  a  transformer,  a  sine  wave  of 
flux  must  be  produced  linking  with  the  primary  winding.  The 
exciting  current  which  produces  the  flux  cannot  be  a  sine  wave 
on  account  of  the  lag  of  flux  due  to  hysteresis.  This  is  shown  in 
Fig.  342.  Curve  (1)  abcf  is  a  hysteresis  loop  for  the  transformer 


(1  )  Hysteresis  Loop 

.6 


(2)  Flux  in  Core 


(3)  Exciting  Current 
f  4)  bine  Wave  Equivalent  to(8) 


FIG.  342. — Exciting  current. 

iron,  plotted  with  values  of  flux  as  ordinates  on  a  base  of  exciting 
current.  Curve  (2)  is  the  sine  wave  of  flux  in  the  core  and  curve 
(3)  is  the  wave  of  exciting  current.  The  method  of  obtaining 
curve  (3)  can  be  seen  from  the  figure.  The  maximum  values  of 
flux  and  current  must  occur  together;  when  the  flux  is  zero  the 
current  has  a  value  oa  or  oc  and  when  the  current  is  zero  the 
flux  has  its  residual  value  od  or  og. 

For  purposes  of  .analysis  the  current  wave  (3)  is  replaced  by 
the  equivalent  sine  wave  (4).  The  current  wave  (4)  leads  the 
flux  wave  (2)  by  an  angle  a,  which  is  called  the  angle  of  hysteretic 
advance.  If  the  eddy-current  loss  is  small  enough  to  be  neglected, 
a  =  90  —  0o,  where  cos  00  is  the  no-load  power  factor. 

304.  Leakage  Reactances. — Figs.  343  and  344  chow  the  leak- 
age paths  around  the  windings  of  a  "shell-type"  and  "core-type" 
transformer.  Since  the  low-voltage  windings  are  placed  next  to 
the  iron,  the  leakage  path  surrounding  the  low-voltage  winding  is 


TRANSFORMERS 


369 


of  slightly  lower  reluctance  than  that  surrounding  the  high-vol- 
tage winding  and  the  reactance  is  correspondingly  larger. 

In  the  shell-type  transformer  the  two  windings  are  divided  into 
a  number  of  sections  and  high-voltage  and  low-voltage  coils 
placed  alternately  to  reduce  the  reactances. 

The  reactance  voltage  of  a  transformer  with  full-load  current 
is  about  10  per  cent,  of  the  total  voltage.  If,  therefore,  full  vol- 


FIG.  343. — Leakage  fluxes  in  a  shell-     FIG.  344. — Leakage  fluxes  in  a  core- 
type  transformer.  type  transformer. 


tage  were  impressed  on  a  transformer  primary  with  the  secondary 
short-circuited  about  ten  times  full-load  current  would  flow. 
When,  however,  full  voltage  is  impressed  on  the  primary  with 
the  secondary  open,  the  exciting  current  which  is  less  than  10 
per  cent,  of  full-load  current  flows.  Thus  the  open-circuit  react- 
ance of  a  transformer  is  of  the  order  of  one  hundred  times  the 
short-circuit  reactance. 

The  vector  diagram  in   Fig.   345   shows  the  various  fluxes 
present  in  a  transformer. 

24 


370  ELECTRICAL  ENGINEERING 

<I>      =  flux  linking  both  primary  and  secondary  windings. 

$L2    =  leakage  flux  surrounding  the  secondary  winding;  it  in- 

duces in  the  secondary  an  e.m.f.  causing  the  secondary 

reactance  drop  I2x2. 
$,     =  total  flux  linking  the  secondary  winding    <f>8  =  $  +  &LZ', 

it  induces  in  the  secondary  an  e.m.f.  E8  in  quadrature 

behind  3>8.     E8  =  $  +  /2r2. 
&LI    =  leakage  flux  surrounding  the  primary  winding;  it  induces 

in  the  primary  an  e.m.f.  causing  the  primary  reactance 

drop  IiXi. 
$>P    =  total  flux  linking   the   primary   winding  <£P  =  <£  -f  $LV 

it  induces  in  the  primary  an  e.m.f.  EP  in  quadrature 

behind  $P. 
E'p  —  component  of  the  impressed  e.m.f.  required  to  overcome 

EP.    £]'p  =  —  EP  =  EI  —  J\TI. 

305.  The  Transformer  as  a  Circuit.  —  The  circuit  in  Fig.  346 
represents  a  transformer  supplying  a  load  of  resistance  R  and 

7") 

reactance  X,  or  of  power  factor  cos  6  =      .  =• 

The  relations  between  the  various  quantities  may  be  expressed 
by  the  following  equations  : 

(333) 

(334) 


W  =  -E*  =  ~h  {(r«  +  R)  +j(x2  +  X)}  (335) 

/fc2  *^2 

/'  =  £!/•  (336) 

/o  =  ^(^0  -  J6o)  (337) 

7i  =  /o  +  /'  (338) 

#1  =  7i?i  -r-  W  (339) 
and  these  equations  may  be  combined  as  follows  : 

Vi  =  Mi  +  W 

=  (/o  +  /'Xn  +  jxd  +  ~/2  (  (rs  +  fl) 


7o(r,  +  j*0  +  /'ri  +          (r2  +  B) 

(340) 


TRANSFORMERS 


371 


This  equation  may  be  represented  by  the  circuit  in  Fig.  347; 
secondary  quantities  are  represented  by  equivalent  primary 
quantities : 

E.m.f.  EZ  in  the  secondary  is  equivalent  to  e.m.f.  E'  =  — Ez 
in  the  primary. 


FIG.  346. — Circuit  diagram  of  a  trans-     FIG.  347. — Equivalent  circuit  diagram 
former.  of  a  transformer. 

E.m.f.  E  in  the  secondary  is  equivalent  to  e.m.f.  —  E  in  the 

72-2 

primary. 

Current  72  in  the  secondary  is  equivalent  to  current  I'  —  — Iz 

in  the  primary. 

Resistance  r2  in  the  secondary  is  equivalent  to  resistance 

")  TZ  in  the  primary. 

Reactance  x2  in  the  secondary  is  equivalent  to   reactance 
(— )  £2  in  the  primary. 

\722/ 


FIG.  348. — Simplified  circuit  diagram  of  a  transformer. 

If  the  exciting  current  is  neglected,  equation  340  becomes 

(341) 

and  the  simplified  circuit  diagram  in  Fig.  348  may  be  used.     The 
resulting  error  is  very  small. 

306.  Examples.  1.  A  stepdown  transformer  with  a  ratio  of.  turns  of 
10  :  1  delivers  100  kw.  at  2,000  volts  to  a  receiver  of  power  factor  80  per 
cent,  lagging.  Determine  the  primary  impressed  voltage,  the  current  and 
the  power  factor. 


372  ELECTRICAL  ENGINEERING 

The  primary  impedance  is 

Zi  =  rl  +  jxl  =  50  +  80;  ohms. 
The  secondary  impedance  is 

?2  =  r2  +  jxz  =  0.6  +  0.8;  ohms. 
The  primary  exciting  admittance  is 

Yo  =  go  -  jb0  =  (2  -  6;)  10~6. 

I  00  r)f)f) 

The  current  output  or  secondary  current  of  the  transformer  is  0  nnn  '    n  OA 

z,UUU  X  O.oU 

=  62.5  and  taking  this  as  axis  the  various  quantities  may  be  expressed  in 
rectangular  coordinates  as  follows: 

Secondary  current 

7  2  =  62.5  +  Oj 
Secondary  terminal  e.m.f. 

E  =  2,000  (cos  6  +  j  sin  0)  =  1,600  +  1,200.;',  where  cos  6  =  0.8 
is  the  load  power  factor. 
Secondary  impedance  e.m.f. 

I,Z2  =  62.5  (0.6  +  0.8.7)  =  37.5  +  50j. 
Secondary  induced  e.m.f. 

$1  =  E  +  h?2  =  1,637.5  +  1,260;. 
Primary  induced  e.m.f. 

Eb  =  E'  =  —  #2  =  10#2  =  16,375  +  12,500j. 

7l2 

Primary  exciting  current 

/0  =  ^'F0  =  (16,375  +  12,500y)  (2  -  6j)  10~6  =0.11  -  0.07;. 
Primary  load  current 


Total  primary  current 

1  1  =  I'  +/  o=6.36  -0.07j. 
Primary  impedance  e.m.f. 

I,Z,  =  (6.36  -  0.07J)  (50  +  8Q/)  =  324  +  505j. 
Primary  impressed  e.m.f. 

#!  =  E'  +  /i£i  =  16,699  +  13,005;. 
Taking  absolute  values, 
primary  impressed  e.m.f. 

E,  =  V(16,6992+(13,005)2  =  21,160  volts, 
primary  current 

/i  =  \/(6^36)2+  (O.'OT)2  =  6.36  amp., 
exciting  current 

/o  =  V  (0.1  1)2  +"(007)  2  =  0.13  amp.. 
primary  induced  e.m.f. 

^'  =  V(16,375)2  +  (lI3~00)2  =  20,60p  volts. 


TRANSFORMERS  373 

The  primary  impressed  e.m.f.  is  inclined  to  the  axis  of  coordinates  at 
an  angle  0',  where 

tan  6'  =   i|^i  =  0.7785  and  therefore  0'  =  37°50'. 
16,o9y 

The  primary  current  is  inclined  to  the  axis  at  an  angle  0",  where 


tan    6"  =  -    ~  =  -  0.011  and  therefore  0"  =  -  0°  40'. 
6.36 

The  angle  of  phase  difference  between  the  primary  current  and  the 
primary  impressed  e.m.f.  is  0i  =  0'  -  0"  =  38°  30',  and  the  primary  power 
factor  is  cos  0i  =  cos  38°  30'  =  0.782,  or  78.2  per  cent. 

The  regulation  of  the  transformer  under  these  conditions  of  loading  is 

2,116  -  2,000 

-  X  100  per  cent.  =  5.8  per  cent. 
z,uuu 

Primary  copper  loss  is 

/iVi  =  (6.36J2  X  50  =  2,020  watts. 
Secondary  copper  loss  is 

/227-2  =  (62.5)  2  X  0.6  =  2,340  watts. 
Iron  loss  is 

E'*g0  =  (20,  600)  2  X  2  X  10~6  =  850  watts. 

The  efficiency  is  therefore 

output 

77  =  —  ^-r-,  -  100  per  cent. 

output  +  losses 

=  roo,6ro?^2io  x  10°  per  cent"  =  95  per  cent" 

2.  If  the  transformer  in  example  1,  with  2,000  volts  impressed,  is  used 
as  a  step-up  transformer  to  charge  a  cable  system  of  negligible  resistance 
and  supplies  5  amp.,  determine  the  secondary  terminal  e.m.f. 

Primary  impedance  is  now  Z\  =  0.6  +  0.8.;. 

Secondary  impedance  is  Z2  =  50  +  8Q/. 

Primary  exciting  admittance  is  Y0'  =  102F0  =  (2  —  6j)10~4. 

Let  the  secondary  terminal  e.m.f.  be  E  and  take  it  as  the  real  axis,  the 
other  e.m.fs.  and  currents  may  then  be  expressed  in  rectangular  coordinates 
Secondary  terminal  e.m.f. 

E  =  E  +  Oj. 
Secondary  current 

/2  =  0  +  5j. 
Secondary  impedance  e.m.f. 

/2£2  =  5j  (50  +  80j)  =  -  400  +  250J. 
Secondary  induced  e.m.f. 

E2  =  $  +  h?2  =  E  -  400  +  250j. 
Primary  induced  e.m.f. 

E'  =  1/10  #2  =  O.IE  -  40  +  25j. 
Primary  load  current 

/'  =  10  72  =  0  +  50j. 


374  ELECTRICAL  ENGINEERING 

Primary  exciting  current 


/</  =  #To'  =  (0.1#  -  40  +  25.7)  (2-6.7)  10~4 

=  { (0.2  E  +  70)  -  (Q.QE  -  290)  jjlQ-4 
Total  primary  current 

Primary  impedance  e.m.f. 

(0.2#  +  70)  -  (0.6#  -  290).7}  10~4  +50./K0.6  +  0.8j) 
(0.6#  -  190)  10~4  -  40}  +  { (  -  Q.2E  +  230)  10~4  +  30}  ]. 

Primary  impressed  e.m.f. 

The  absolute  value  of  primary  impressed  e.m.f.  is 
E, 


#-  80)2  +  (55)2=  2,000 
and  solving  this  gives  the  secondary  terminal  e.m.f. 
E  =  20,800. 

307.  Measurement  of  the  Constants  of  a  Transformer.  —  The 

constants  are:  (1)  the  ratio  of  turns  —,  (2)  the  primary  exciting 


admittance  Fo  =  "\/<7o2  +  &o2,  (3)  the  primary  impedance  Zi  — 
\A*i2  +  Xi2  and  (4)  the  secondary  impedance  Zz  =  \/^*22  +  ^22. 
The  first  two  can  be  obtained  from  an  open-circuit  test  and  the 
last  two^from  a  short-circuit  test. 


Ei-E 


E2-E 


Fio.  349. — Open-circuit  test. 


FIG.  350. — Short-circuit  test. 


For  the  open-circuit  test,  Fig.  349,  impress  full  voltage  EI  on 
the  primary,  at  rated  frequency,  and  read  primary  impressed 
voltage  Eif  secondary  terminal  voltage  E  =  E2)  primary  exciting 
current  Jo  and  the  power  input  Wo  which  gives  the  core  loss  plus 
a  very  small  copper  loss  IQ^TI  which  may  be  neglected. 

The  ratio  of  turns  is  found  very  approximately  as  —  =     -  = 


Tjl 


T 

w  > 


p-;  the  primary  exciting  admittance  is  Y0  =  VfToM-  &o 

TFo  TFo 

the  core-loss  current  is  7C  =    ™—  =  Eig0,   and  thus,   <7o   =    jjn', 


TRANSFORMERS  375 


the  magnetizing  current  is  IM  =  V/o2  —  Ic2  =  Eib0,  or  b0  may  be 
found  as  60  =  X/F02  —  go2',  the  no  load  power  factor  is  cos  0o  = 

^-j-  and  the  angle  of  hysteretic  advance  is  a  =  90  —  0o,  if  the 

&llQ 

eddy-current  loss  is  neglected. 

In  the  short-circuit  test,  Fig.  350,  the  secondary  is  short- 
circuited  and  a  voltage  Esc  is  impressed  on  the  primary  of  such  a 
value  that  full-load  current  /i  flows;  the  secondary  will  then  carry 
full-load  current  1 2',  readings  are  taken  of  E8C,  Ii  and  the  power 

W 
input  W8C:  the  power  factor  is  found  as  cos  68C  =  ^  8l  •   Theex- 

XWi 

citing  current  is  so  small  that  it  may  be  neglected. 

Since  the  terminal  voltage  is  zero,  the  impressed  voltage  Etc 
is  consumed  by  the  impedance  of  the  transformer  and  is  the  full- 
load  impedance  drop.  When  expressed  as  a  per  cent,  of  the 
rated  primary  voltage,  it  is  the  per  cent,  impedance  drop.  The 
per  cent,  resistance  drop  can  be  found  by  multiplying  by  cos  68C 
and  the  per  cent,  reactance  drop  by  multiplying  by  sin  0,c. 

Esc     —    IlZi   -\ /2^2 


-•+©"  )'+('•+©'* 


where  r,  x  and  Z  are  the  equivalent  resistance,  reactance  and  im- 
pedance of  the  transformer  as  primary  quantities. 

The  power  input  W8C  =  primary  and  secondary  copper  losses 
plus  a  very  small  core  loss,  which  may  be  neglected. 


W8C  =  7M  +  /2V2  =  IS    n  + 

and  thus 

W8C 


to  determine  the  values  of  TI  and  r2,  it  may  be  assumed  that  the 
two  windings  have  been  designed  for  the  same  current  density 
and  that  therefore  the  two  copper  losses  are  approximately  the 
same  and 


376 


ELECTRICAL  ENGINEERING 


In  a  similar  way  x  =  Xi  +  (— )  x2  may  be  separated  into  i 

XTio/ 


its 


two  components  x\  =    ~     x2  =  ^'     This  is  correct  if  the  leakage 

fluxes  about  the  two  windings  are  equal,  which  may  be  assumed 
to  be  the  case,  since  the  m.m.fs.  are  equal  and  the  leakage  paths 
are  very  similar. 

308.  Regulation.  —  The  regulation  of  a  transformer  is  the  rise 
in  secondary  terminal  voltage,  when  full  load  is  thrown  off,  ex- 
pressed as  a  per  cent,  of  full-load  voltage.  The  regulation  of  a 
transformer  is  very  much  better  than  that  of  an  alternator  be- 
cause the  resistance  and  reactance  drops  are  very  much  smaller. 

The  regulation  at  100  per  cent,  power  factor  ranges  from  1.2 
to  2  per  cent,  for  large  25-cycle  transformers  and  from  0.75  to  1.6 
per  cent,  for  large  60-cycle  power  transformers.  Larger  values 
are  found  in  some  small  transformers.  The  regulation  of  large 
power  transformers  may  be  as  high  as  6  or  7  per  cent  at  80  per 
cent,  power  factor. 

If  the  terminal  voltage  at  full  load  is  E  =  100  per  cent,  and  the 
impedance  drop  is  expressed  in  per  cent.,  the  vector  sum  of  the 
two  will  give  the  terminal  voltage  at  no  load  in  per  cent.  Taking 


. 

s, 

y 

2 

3     4 

0    0 

o  a 

1  * 

xKso  c 

T  — 

0    4 

»    J 

o 

C 

^ 

^ 

Lag 


Lead 


Per  Cent  Power  Factor 


FIG.  351. — Regulation  of  a  transformer. 


the  current  as  axis  and  a  load  power  factor  cos  6,  the  secondary 
terminal  voltage  is  E  =  100  cos  8  -fjlOOsinfl;  the  impedance 
drop  is  IZ  =  Ir  +  jlx,  all  in  per  cent.,  and  the  terminal  voltage 
at  no  load  is  E<>  =  E  +  IZ  =  (100  cos  6  +  Ir)  +  j(lQQ  sin  0  + 
Ix);  its  absolute  value  is 


E0  =  A/(100  cos  B  +  Ir)2  +  (100  sin  6  +  Ix)2        (342) 

and  the  per  cent,  regulation  is  E0  —  100. 

A  formula  giving  the  regulation  directly  may  be  found  by  re- 
ferring to  the  vector  diagram  in  Fig.  351  (a).     The  per  cent. 


TRANSFORMERS  377 

regulation  for  a  load  power  factor  cos  6  is  represented  by  the 
length  ad  = 

D  =  ab  +  be  +  cd  =  Ir  cos  0  +  /z  sin  6  +  cd. 


From  Pig.  351  (6),  cd  X  ck  =  7f  and  cd  =        = 

(Ix  cos  0  -  Jr  sin  BY       ,    .  .  4. 

—  ^  ,   ,  —      -  and  the  per  cent,  regulation  is 

T  T     •         ,  '(!%  cos  0  —  Ir  sin  0)2       ,0  ._. 

D  =  7r  cos  0  +  Is  sin  0  +  —  (343) 


where  Ir  is  the  per  cent,  resistance  drop  and  Ix  the  per  cent,  re- 
actance drop  in  the  transformer. 

The  last  term  is  negligible  except  near  unity  power  factor 
and  if  it  is  neglected,  the  per  cent,  regulation  is 

D  =  Ir  cos  6  +  Ix  sin  6  =  IZ  (^  cos  6  +  ~  sin  d\ 

=  IZ  (sin  0  cos  e  +  cos  <f>  sin  6)  =  IZ  sin  (0  +  <£)  ;  (344) 

\  r 

the  angle  <£  =  tan-1  -  is  shown  in  Fig.  351. 
x 

The  regulation  is  maximum  when  sin  (0+  <£)  =  1,  or  6  =  90  — 
</>,  that  is,  at  a  load  power  factor  cos  6  =  cos  (90  —  0)  =  sin  <f>  = 

-y  and  is  equal  to  the  per  cent,  impedance  drop  IZ. 

Take  for  example  a  single-phase,  60-cycle,  33,  000-  volt  trans- 
former of  100  kva.  output  with  a  resistance  drop  of  1.5  per  cent. 
and  a  reactance  drop  of  4  per  cent.,  and  calculate  the  regulation 
at  100  per  cent,  power  factor  and  80  per  cent,  power  factor  lag 
and  lead,  using  equation  (343)  : 

At  100  per  cent,  power  factor,  the  regulation 

=  1.5  +  ^  =  1.5  +  0.08  =  1.58  per  cent. 
At  80  per  cent,  power  factor  lag  the  regulation 


=  1.2  +  2.4  +  0.026  =  3.626  per  cent. 
At  80  per  cent,  power  factor  lead,  the  regulation 

=  1.5  X  0.8  -  4  X  0.6  +  (IX2 


=  1.2  -  2.4  +  0.08  =  -  1.12  per  cent. 


378 


ELECTRICAL  ENGINEERING 


50  Percent  Power  Factor  Leadi 


80  Percent  Power  Factor 

ower  Factor  Lagging 


aggi 


Amperes  Load  Current   I 

FIG.  352. — Voltage  characteristics  of  a 
transformer. 


The    maximum    regulation   is  Z>max.  =  IZ  =  vlr2  +  Ix2  — 
*\/1.52  +  42  =  4.27  per  cent,  and  corresponds  to  a  load  power- 

T      i  5 

factor,  cos  6  =  -^  =  7-5=  =  0.35  =  35  per  cent. 

The  complete  regulation  curve  for  this  transformer  is  shown 
in  Fig.  351(c).  For  leading  power  factors  the  regulation  becomes 

negative,  that  is,  the  voltage 
rises  with  load. 

309.  Voltage  Characteris- 
tics.— The  voltage  character- 
istics of  a  transformer  show 
the  relation  between  the 
secondary  terminal  voltage 
and  the  secondary  load  cur- 
rent for  given  values  of  power 
factor,  with  a  constant  im- 
pressed voltage  Ei.  Typical 
voltage  characteristics  are 
shown  in  Fig.  352  plotted  in 
per  cent.  Since  the  no-load 

secondary  voltage  is  constant  =  EQ  =  —Ei,  it  has  been  taken  as 

100  per  cent,  and  the  values  of  E  calculated  from  the  equation, 

EQ  =  100  =  V(E  cos  e  +  /r)2  +  (E  sin  0  +  Ix)2. 

310.  Losses  in  Transformers. — The  losses  in  a  transformer 
are  the  copper  losses  and  the  iron  losses. 

The  total  copper  loss  is 

Wc  =  I&i  +  72V2  (345) 

and  is  almost  evenly  divided  between  the  primary  and  the  second- 
ary since  the  two  windings  are  designed  for  approximately  the 
same  current  density. 

The  iron  loss  consists  of  the  hysteresis  and  eddy-current  losses 
and  does  not  vary  with  load.  It  can  be  measured  by  applying 
full  voltage  to  the  primary  with  the  secondary  open  and  reading 
the  watts  input,  r  **\ 

W o  =  EJQ  cos  00  =  Eilc  =  Ei*gQ.  ^346) 

Wo  includes  a  small  copper  loss  /oVi  which  may  be  neglected. 

311.  Hysteresis  Loss. — The  hysteresis  loss  is  the  energy  con- 
sumed in  magnetizing  and  demagnetizing  the  iron.     It  is  directly 


TRANSFORMERS 


379 


proportional  to  the  frequency/,  varies  as  the  1.6th  power  of  the 
maximum  induction  density  <Bm,  and  also  depends  on  the  quality 
of  the  iron. 

The  hysteresis  loss  per  cycle  per  cubic  centimeter  of  iron  is 

(347) 


wh  = 


ergs, 


where  17  is  the  hysteresis  constant  for  the  iron  and  has  a  value  of 
about  0.003  for  good  transformer  punchings.  In  silicon-steel 
plates  rj  may  be  as  low  as  0.001. 

The  hysteresis  loss  per  second  in  a  volume  of  V  c.c.  at  a  fre- 
quency /  is 

Wh  =  yfV($>ml'&  ergs  per  second  (348) 

=  rjfV(8>ml-«  10-7  watts. 

At  constant  frequency  the  hysteresis  loss  varies  approximately 
as  the  1.6th  power  of  the  impressed  voltage. 

312.  Eddy-current  Loss  in  Transformer  Iron.  —  The  eddy- 
current  loss  is  the  energy  consumed  by  the  currents  induced  in 
the  iron  of  the  transformer  by  the  alternating  flux  cutting  it. 

In  Fig.  353 

t  =  thickness  of  sheets  in  centimeters, 
($>„>  =  maximum  induction  density, 
/  =  frequency, 
7  =  electric  conductivity  of  iron. 

AB  is  a  section  of  length  1  cm.,  depth  1 
cm.  and  width  dx  cm.,  parallel  to  the  edge 
of  the  plate  and  distant  x  cm.  from  the 
center.  DC  is  a  similar  section  on  the 
other  side. 

The.  maximum  'flux  inclosed  by  ABODA9>2($>mx  lines.  Of  this 
(&mx  lines  cut  across  the  section  AB  and  generate  in  it  an  e.m.f.  of 
effective  value, 

dE  =  \/2Trf($>mx  c.g.s.  units. 

The  conductance  of  the  section  A  B  is  7  dx,  and  thus  the  current 
induced  in  it  is 

dl  =  dEy  dx  =  \/2irf($>myx  dx  c.g.s.  units. 
The  power  consumed  by  the  induced  current  in  section  is 
dwe  =  dE  dl  =  2Tr2f2(&m2yx2  dx  ergs  per  second 


-*-&- 

1 

—  H- 

A"* 

kr*> 

dx 

t 

i 

i 
1 

c 

i 

B 

FIG.  353. — Eddy  current 
loss  in  sheet  iron. 


380  ELECTRICAL  ENGINEERING 

and  thus  the  power  consumed  per  square  centimeter  of  the  plate 
is 


=  ir1/2®*2  7  TT  ergs  per  second.  (349) 

The  power  consumed  per  cubic  centimeter  of  iron  is 


t  D 

Thus  the  eddy-current  loss  for  a  volume  of  V  c.c.  is 


ITT  "  m     f  •     '  i 

Tr  e  =  -     —  ^—     —  ergs  per  second 

=  ^2(V^2F  10-7  watts.  (351) 

D 

The  eddy-current  loss,  therefore,  varies  as  the  square  of  the 
frequency,  the  square  of  the  maximum  induction  density  and  the 
square  of  the  thickness  of  the  plates.  It  also  varies  directly  with 
the  electric  conductivity  of  the  iron. 

To  reduce  the  eddy-current  loss  the  core  is  usually  built  up  of 
sheets  of  about  0.014  in.  thick  insulated  with  varnish.  The 
electric  conductivity  of  ordinary  sheet  iron  is  about  105.  Silicon 
steel  has  a  conductivity  in  some  cases  as  low  as  2  X  104  and  it 
may  be  used  in  sheets  0.020  in.  thick. 

With  a  sine  wave  of  impressed  voltage  the  eddy-current  loss 
varies  as  the  square  of  the  voltage,  but  if  the  wave  is  peaked  the  ^ 
loss  may  be  greater,  {  p**^*  &  <**&*£»  %*4*>*  j  {£4  ^  Jb^U 

Since  iron  has  a  positive  temperature  coefficient  for  resistance  x 
the  eddy-current  loss  will  decrease  slightly  as  the  transformer 
heats  up. 

313.  Extra  Losses.  —  When  the  windings  of  a  transformer  are 
made  up  of  a  number  of  sections  connected  in  multiple,  if  the 
impedances  of  the  sections  are  not  the  same,  circulating  currents 
will  flow  and  increase  the  copper  losses.  They  will  exist  at  no 
•load  as  well  as  under  load. 

If  wide  copper  strip  is  used  for  the  winding,  eddy  currents  may 
flow  in  the  copper  due  to  the  fact  that  one  side  is  in  a  stronger 


TRANSFORMERS  381 

field  than  the  other.  This  may  increase  the  copper  loss  as  much 
as  20  per  cent,  in  extreme  cases.  It  is  not  advisable  to  use  strip 
of  width  greater  than  0.50  in.  in  60-cycle  transformers  or  greater 
than  0.75  in.  in  25-cycle  transformers. 

314.  Efficiency.  —  The  efficiency  of  a  transformer  is 

output 

r;  =  -  —  —,  —       —  :  --  ;  --  100  per  cent. 

output  -f  copper  loss  +  iron  loss 

EI2  cos  9 
=  -  -  100  per  cent. 

EI2  cos  e  +  I\rl  +  7fr2  +  k 
=  -  EI2  CosO 
EI2  cos  e  +  l\r  +  k 

where  r  =  (—  )    r\  +  r2  is  the  equivalent  secondary  resistance 

\Yl\J 

and  k  =  constant  iron  loss. 

The  efficiency  is  very  high  and  remains  high  over  a  wide  range 
of  load;  it  reaches  its  maximum  value,  for  a  given  load  power 
factor,  when  the  variable  copper  loss  =  the  constant  iron  loss. 
This  may  be  shown  by  differentiating  rj  with  respect  to  Iz  and 
equating  the  result  to  zero. 

^L 
dI2 

{  (EI2  cos  8+I22r+k)E  cos  6-EI2  cos  8  (E  cos  6  +  2J2r)  1 

(#J2cos0  +  /22r  +  /c)2 
for  maximum  efficiency. 

Therefore,  E  cos  0  (EI2  cos  0  +  IJr  +  k  -  EI2  cos  9  -  2I22r) 
=  0  and  I22r  =  k,  that  is,  the  copper  loss  =  the  iron  loss.  Below 
one-fourth  load  the  efficiency  falls  off  rapidly  due  to  the  constant 
iron  losses  and  above  full  load  it  falls  due  to  the  large  copper 
losses. 

The  efficiency  also  depends  on  the  power  factor  of  the  load, 
decreasing  with  it  but  not  in  direct  proportion. 

In  the  smaller  sizes  the  efficiency  at  unity  power  factor  varies 
from  93  per  cent,  at  one-fourth  load  to  97  per  cent,  at  full  load 
and  in  the  larger  sizes  from  97  per  cent,  at  one-fourth  load  to 
99  per  cent,  at  full  load. 

Take  for  example  a  transformer  rated  at  100  kva.  with  a  full- 
load  copper  loss  of  2,000  watts  and  an  iron  loss  of  1,600  watts. 

The  efficiency  at  full-load  unity  power  factor  is 

100,000 


100,000  +  2,00  + 


100per  cent>  =  96'5  per  cent'; 


382 


ELECTRICAL  ENGINEERING 


the  efficiency  at  full-load  80  per  cent,  power  factor  is 

100,000  X  0.80 
77  =  100,000  X  0.80  +  2,000  +^600  X<  I0per  Cent'  =  9 

the  efficiency  at  one-fourth  load  unity  power  factor  is 


100,000 


000 


=   93'6 


Cent' 


1,600 


In  the  case  of  distributing  transformers  which  are  connected 
to  the  supply  lines  at  all  times  but  are  delivering  power  for  only  a 
few  hours  during  the  day,  the  all-day  efficiency  is  of  more  impor- 
tance than  the  actual  efficiency.  It  is  the  ratio  of  the  energy  out- 
put during  the  day  to  the  energy  input. 


50  100 

Percent  Load 

FIG.  354.  —  Efficiency  and  loss  curves  of  a  550-kw.,  10,500-volt,  60-cycle,  air- 

blast  transformer. 

If  the  transformer  mentioned  above  is  operated  at  full-load 
unity  power  factor  for  6  hr.  during  the  day  and  is  unloaded  the 
remainder  of  the  time,  the  all-day  efficiency  is 

,,   ,  energy  output  .,__ 

17  all-day  =  -    -^—  —     -  100  per  cent. 
energy  input 

100,000X6 


100,000X6+2,000X6+1,600X24 
When  a  good  all-day  efficiency  is  required,  the  iron  losses  should 
be  kept  small  by  using  the  best  grades  of  iron  or  by  reducing  the 
flux  density. 

Fig.  354  shows  efficiency  and  loss  curves  for  a  large  transformer. 


TRANSFORMERS 


383 


315.  Types  of  Transformers. — Transformers  may  be  divided 
into  two  general  types  depending  on  the  arrangement  of  the  core 
and  the  windings:  the  core  type,  (Fig.  355);  and  the  shell  type 
(Fig.  356). 

In  the  core  type  a  single  ring  of  iron  is  surrounded  by  two 
groups  of  windings,  while  in  the  shell  type  a  single  ring  of  copper 
formed  of  the  two  windings  is  surrounded  by  two  or  more  rings 
of  iron.  The  core  type  has  a  long  magnetic  path  and  a  short 


nil  ill  ill  |  ill!  I!,*  | 


FIG.  355. — Core-type  transformer.      FIG.  357. — Three-phase    core-type    trans- 
formers. 


Unit  of  Equivalent  Single 
Phase  Group 


FIG.  356. — Shell-type  transformer. 


FIG.  358. — Three-phase    shell-type     trans- 
former. 


mean  turn  of  winding  and  the  shell  type  has  a  short  magnetic 
circuit  and  a  long  electric  circuit.  The  core  type  is  more  suitable 
for  small  high-voltage  transformers  as  it  is  very  easy  to  insulate 
the  high-voltage  from  the  low-voltage  coils  by  placing  a  cylinder 
of  insulating  material  between  them.  Large  power  transformers 
are  usually  made  of  the  shell  type  but  core-type  transformers  can 
be  designed  with  equally  good  characteristics. 

Three-phase  transformers  are  built  of  both  the  core  and  the 
shell  types  (Figs.  357  and  358) .     In  Fig.  358  a  three-phase  trans- 


384  ELECTRICAL  ENGINEERING 

former  is  compared  with  one  of  the  equivalent  group  of  single- 
phase  transformers.  Since  the  fluxes  in  the  three  phases  are 
combined  at  120  degrees  in  the  common  paths  a  considerable 
saving  in  core  material  is  possible. 

In  the  majority  of  polyphase  systems  groups  of  single-phase 
transformers  are  used  instead  of  three-phase  units  but  the  three- 
phase  units  have  some  advantages  which  tend  to  increase  their 
application. 

A  three-phase  transformer  is  less  expensive  than  the  equivalent 
single-phase  group  requiring  a  smaller  weight  of  iron,  a  single 
tank,  and  fewer  high-voltage  brushings,  since  the  connections 
between  phases  are  made  inside  the  tank.  The  floor  space  re- 
quired is  very  much  less  and  the  bus-bar  and  switching  layouts 
are  simplified,  as  is  also  the  cooling  water  system.  A  spare  three- 
phase  unit  must  be  installed  instead  of  a  spare  single-phase  unit 
but  the  extra  capacity  may  be  very  valuable  at  times  of  overload. 
For  very  large  outputs  three-phase  units  may  be  too  heavy  to 
handle  easily  in  case  repairs  are  necessary  and  single-phase  units 
may  be  more  satisfactory. 

If  one  phase  of  a  three-phase  delta-connected  transformer  is 
damaged,  operation  may  be  carried  on  with  reduced  output 
using  the  other  two  windings  in  open-delta  but  both  the  primary 
and  secondary  windings  of  the  damaged  phase  must  be  discon- 
nected and  short-circuited. 

316.  Methods  of  Cooling. — Very  small  transformers  do  not 
require  any  special  method  of  cooling  but  are  so  designed  that  the 
exposed  surface  is  large  enough  to  radiate  the  heat  generated  by 
the  power  losses  in  the  windings  and  core  without  a  temperature 
rise  exceeding  the  limits  consistent  with  the  life  of  the  insula- 
tion. Transformers  must  be  designed  to  operate  for  24  hr.  at 
full  load  with  a  temperature  rise  not  exceeding  55°C.  above  the 
ambient  temperature  of  40°C. 

Since  the  output  and  losses  in  a  transformer  increase  in  pro- 
portion to  its  volume  or  as  the  cube  of  its  linear  dimensions 
while  the  radiating  surface  increases  only  as  the  square  of  the 
linear  dimensions,  as  the  output  is  increased  special  methods  of 
cooling  must  be  adopted. 

Transformers  up  to  500  kw.  are  usually  immersed  in  tanks 
containing  oil  of  good,  insulating  qualities.  This  oil  serves  the 
double  purpose  of  increasing  the  insulation  of  the  transformer  and 
conducting  away  the  heat  developed  by  the  losses.  Such  trans- 


TRANSFORMERS 


385 


formers  are  called  oil-insulated  self-cooled  transformers.  The 
cases  are  made  with  deep  corrugations  to  give  a  larger  radiating 
surface  exposed  to  the  air. 

The  oil,  as  it  comes  into  contact 
with  the  coils  and  core,  becomes 
heated,  rises  to  the  top  of  the  tank 
and  flows  outward  to  the  sides  where 
it  gives  up  its  heat  to  the  air  and  falls 
again  to  the  bottom  of  the  tank.  A 
continuous  circulation  of  oil  is  thus 
maintained  and  care  must  be  taken 
that  paths  of  ample  section  are  left 
for  the  passage  of  the  oil  through  the 
transformer  (Fig.  359). 

Where  larger  self-cooled  units  are 
required,  as  in  the  case  of  large  out- 
door substances,  special  methods  of 
design  are  necessary  to  provide  sufficient  radiating  surface.  Two 
successful  types  are  shown  in  Fig.  360,  the  tubular  tank  and 
the  external  radiator  tank. 

The  tubular  tank  has  a  large  number  of  tubes  welded  to  it, 
which  draw  the  hot  oil  from  the  top  of  the  tank  and  feed  it  back 


( 

Ml  ] 

LeVe 

1 

s 

•^N 

^ 

f 

Y 
1 

Cc 

il 

X 
i 

J 

I 

1 

1 

I 
t 
t 

L 

»/- 

t 
t 
t 
tj 

|E 

I 
1 
1 
j 

( 

^ 

1 

r 

C 

/ 

are 

\ 

< 

Y 

y 

adiator 


Tubular  tank.  External   radiator    tank. 

FIG.  360. — Oil-insulated  self-cooled  transformers. 

near  the  bottom  after  its  temperature  has  been  reduced.  The 
tubes  are  very  efficient  radiators  and  the  effective  surface  of  the 
tank  may  be  increased  many  times  by  their  use.  Self-cooled 
units  up  to  2,000  kva.  may  be  built  in  this  way. 

25 


386  ELECTRICAL  ENGINEERING 

A  second  method  of  increasing  the  radiating  surface  of  the 
tank  is  shown  in  Fig.  360.  Radiators  made  of  corrugated  iron 
with  very  large  surfaces  are  used  to  replace  the  tubes  in  the  last 
design.  They  are  made  so  that  they  can  be  easily  detached 
when  the  transformer  is  to  be  shipped.  Units  of  this  type  with 
an  output  of  5,000  kva.  are  in  successful  operation. 

These  special  tanks  are  very  bulky  and  are  not  suitable  for 
installation  where  space  is  limited. 

In  the  case  of  large  banks  of  transformers  it  may  be  econom- 
ical to  provide  a  forced  oil  circulation.  The  hot  oil  is  pumped 
from  the  transformers  to  a  chamber  where  it  is  cooled  by  an  air 
blast  or  in  some  other  way. 

A  second  method  of  getting  rid  of  the  heat  developed  by  the 
losses  is  to  blow  air  through  the  transformers  to  cool  them. 
Transformers  cooled  in  this  way  are  termed  "air-blast"  and  are 
used  on  electric  locomotives  where  their  light  weight  is  an  advan- 
tage and  in  places  where  oil  cannot  be  used  on  account  of  the  dan- 
ger of  fire.  They  cannot  be  operated  safely  above  about  30,000 
volts  as  the  insulation  rapidly  deteriorates  due  to  ozone  set  free 
in  the  air  at  high  voltages.  Oil  is  then  a  necessary  protection. 

For  air-blast  transformers  about  150  cu.  ft.  of  air  are  required 
per  minute  per  kilowatt  lost  to  keep  the  temperature  within  the 
permissible  limits. 

The  majority  of  large  transformers  are  cooled  by  placing  in 
the  upper  part  of  the  tank  cooling  coils  which  carry  a  continuous 
flow  of  cold  water,  which  conveys  the  heat  away  from  the  oil. 
With  the  incoming  water  at  25°C.,  about  Y±  gal.  per  minute  pe- 
kilowatt  lost  is  required.  If  too  much  water  is  supplied  the  trans- 
former may  be  cooled  below  the  temperature  of  .the  air  and  mois- 
ture may  collect  on  it  and  cause  a  breakdown  of  the  insulation. 
The  cooling  coils  should  have  a  surface  in  contact  with  the  oil  of 
about  1.0  sq.  in.  per  watt  lost. 

Impurities  in  the  water  are  liable  to  collect  in  the  pipes  and 
interfere  with  the  flow. 

Combined  self-cooling  and  water-cooled  transformers  are 
sometimes  built.  The  radiating  surface  of  the  tank  is  large 
enough  to  take  care  of  the  losses  up  to  half  load  but  above  this 
point  water  must  be  circulated  through  the  coils  to  keep  down  the 
temperature. 

317.  Transformer  Connections. — If  the  primary  and  sec- 
ondary windings  are  each  divided  into  a  number  of  coils,  which 


TRANSFORMERS 


387 


can  be  connected  either  in  series  or  parallel,  a  number  of  differ- 
ent ratios  of  transformation  can  be  obtained. 

Take  for  example  a  standard  lighting  transformer  with  two  coils 
for  110  volts  on  the  low-voltage  side  and  two  coils  for  1,100  volts 


Ratio  2200:220=10:1          Ratio  2200:110=20:1  Ratio  1100:220=5:1 

FIG.  361. — Transformer  connections. 


tRatio  1100:110=10:1 


on  the  high-voltage  side.     The  four  possible  connections  are 
shown  in  Fig.  361. 

If  small  percentage  changes  01  ratio  are  required  for  line  regula- 
tion a  number  of  taps  are  brought  out  from  the  primary  or 
secondary  winding  so  that  the  number 
of  turns  in  use  may  be  changed  by  the 
required  amount  (Fig.  362). 

The  secondary  voltage  may  be  in- 
creased by  cutting  out  some  of  the  pri- 
mary turns  or  it  may  be  decreased  by 
cutting  out  some  of  the  secondary  turns. 

318.   Single -phase    Transformers    on 
Polyphase  Circuits. — Single-phase  trans- 
formers are  used  in  groups  on  polyphase  circuits.     The  principal 
transformations    are    two-phase   to   two-phase,    three-phase   to 
three-phase,   two-phase  to  three-phase  or  three-phase  to  two- 
phase,  and  three-phase  to  six-phase. 


FIG.  362. — Variable  ratio 
transformer. 


Star- Star  Delta-Star  Star- Delta  Delta-Delta 

FIG.  363. — Single-phase  transformers  on  three-phase  circuits. 

The  transformation  from  two-phase  to  two-phase  needs  no 
discussion.  The  two  phases  are  usually  entirely  separate  but 
they  may  be  interconnected  if  required. 

Transformation    from    three-phase    to    three-phase    can    be 


388  ELECTRICAL  ENGINEERING 

carried  out  in  a  large  number  of  ways,  the  most  important  of 
which  are,  star-star,  delta-star,  star-delta,  delta-delta,  open- 
delta  and  tee  (Figs.  363,  366  and  367). 

Transformation  from  two-phase  to  three-phase  is  obtained  by 
the  use  of  the  Scott  connection,  Fig.  369,  and  is  reversible. 

Six-phase  power  is  required  for  the  operation  of  many  rotary 
converters.  It  can  be  obtained  from  a  three-phase  circuit  by 
the  double-delta  connection,  the  diametrical  or  double-star 
connection  and  the  ring  connection,  Fig.  370.  Other  methods 
of  connection  are  possible  such  as  the  double  tee.  Six-phase 
power  can  be  obtained  from  two-phase  circuits  by  using  two 
Scott  connected  banks  of  transformers. 

319.  Star-star  Connection. — The  star-star  connection  may  be 
used  to  transform  three-phase  power  to  three-phase  power  at 
another  voltage  but  it  has  some  disadvantages  which  restrict  its 
use  to  special  cases. 

In  Art.  303  it  was  shown  that,  when  a  sine  wave  of  voltage  is 
impressed  on  a  transformer  and  a  sine  wave  of  flux  is  produced  in 
the  core,  the  exciting  current  is  not  a  sine  wave  but  has  the  shape 
shown  in  Fig.  342.  When  such  a  wave  is  analyzed  as  in  Art. 
121  it  is  found  to  consist  of  a  fundamental  sine  wave  with  a  very 
prominent  third  harmonic  and  higher  harmonics  of  smaller 
amplitude* 

The  fundamental  sine  wave  of  exciting  current  or  the  equiva- 
lent sine  wave  in  Fig.  342  may  be  separated  into  two  sine  waves 
as  in  Fig.  364  the  core-loss  current,  curve  3,  in  phase  with  the 
impressed  e.m.f.,  curve  1,  and  the  magnetizing  current,  curve  4, 
in  quadrature  behind  the  impressed  e.m.f.  and  in  phase  with  the 
flux,  curve  2.  Due  to  saturation  a  third  harmonic  magnetizing 
current,  curve  5,  must  be  added  to  the  fundamental  to  produce 
the  sine  wave  of  flux.  The  shape  of  the  resultant  magnetizing 
current  wave,  curve  6,  is  therefore  symmetrical  with  a  peak  at  the 
center.  The  wave  of  exciting  current,  curve  7,  is  found  as  the 
sum  of  3  and  6,  and  has  the  shape  shown  in  Figs.  342  and  364. 
The  amplitude  of  the  third  harmonic  increases  with  the  degree  of 
saturation  of  the  core. 

In  the  three  branches  of  the  star  connection,  the  three  funda- 
mental currents  are  displaced  at  120  degrees  while  the  third 
harmonic  currents  are  displaced  120  X  3  =  360  degrees  and  are 
therefore  in  phase.  For  the  third  harmonics  the  ungrounded 
three-phase  circuit  is  like  a  single-phase  circuit  without  any 


TRANSFORMERS 


389 


return.  When  the  neutral  point  is  not  grounded  the  third  har- 
monic and  its  multiples  in  the  exciting  current  cannot  flow  and 
are  therefore  suppressed. 

The  result  is  that  the  flux  wave  is  no  longer  a  sine  wave  but 
may  be  considered  to  be  made  up  of  a  sine  wave,  curve  3,  Fig. 
365,  and  a  third  harmonic,  curve  4.  The  third  harmonic  flux 
is  opposite  in  phase  to  the  missing  third  harmonic  of  magnetizing 
current  and  the  resultant  flux  wave,  curve  2,  is  flat  on  the  top 
and  has  very  steep  sides. 

The  sine  wave  of 'flux  induces  a  sine  wave  of  voltage,  curve  8, 
in  the  two  windings,  in  quadrature  behind  itself  and  the  third 


(1)= Impressed  Voltage 

(2)  =  Flux 

(3)=  Core  Lpss  Current 

(  4)  =  Fundamental  Wave  of  Magnetizing  Current 

(  5  ) = Third  Harmonic-  Magnetizing  Current 

(C)=  Resultant  Magnetizing  (Jurrent 

(7)=Eiciting  Current 


FIG.  364. 


l)=Impressed  Voltage 

Flux 

( 3)= Fundamental  Ware  of  Flux 
(4;=Third  Harmonic  Flux 
(6)=Core  Loss  Current 
(G)=Magnetizing  Current 
(7)=£xclting  C«rrent 
(8)=Fundamental  Induced  Voltage 
(9)=Thlrd  Harmonic  Induced 

Voltage 
(10)=Resultant  Induced  Voltage 


FIG.  365. 


harmonic  flux  induces  a  third  harmonic  voltage,  curve  9,  in 
quadrature  behind  itself.  The  two  components  of  the  induced 
voltage  combine  to  form  the  peaked  wave,  curve  10. 

The  maximum  value  of  the  third  harmonic  voltage  induced  in 
the  secondary  is  2?r  X  3/n2$3 10~8  =  67r/n2$3  10~8  volts,  where  3>3 
is  the  maximum  value  of  the  third  harmonic  flux.  Since  the 
secondaries  are  star-connected,  the  third  harmonic  cannot  appear 
in  the  voltage  between  lines  but  it  will  increase  the  maximum 
value  of  the  voltage  in  each  transformer  and  may  cause  a  break- 
down of  the  insulation.  The  effective  value  of  the  secondary 
voltage  may  be  increased  only  5  or  10  per  cent.,  while  the  maxi- 
mum value  is  increased  40  or  50  per  cent.  This  is  a  very  dan- 


390  ELECTRICAL  ENGINEERING 

geroiis  condition  in  high-voltage  transformers  where  the  factor 
of  safety  is  small. 

If  the  primary  neutral  is  connected  to  the  generator  neutral  the 
third  harmonic  of  exciting  current  can  flow,  returning  along  the 
neutral  and  there  will  be  no  distortion  of  the  voltage  wave. 

If  the  neutral  on  the  secondary  side  only  at  the  generating  end 
of  a  transmission  line  is  grounded  and  the  neutral  on  the  primary 
side  at  the  receiving  end,  the  third  harmonics  in  the  induced 
secondary  voltage  waves  will  cause  third  harmonic  currents  to 
flow  in  the  three  lines  and  since  they  are  in  phase  with  one  another 
they  may  produce  inductive  disturbances  in  neighboring  tele- 
phone lines. 

The  star-star  connection  cannot  be  used  to  supply  a  three- 
phase  four-wire  system  since  the  three  outers  pulsate  against  the 
neutral  connection  at  triple-frequency. 

In  star-star  connected  three-phase  core-type  transformers,  on 
account  of  the  mutual  inductance  between  phases  the  third 
harmonic  in  the  induced  voltage  wave  is  suppressed. 

Another  objection  to  the  use  of  this  connection  is  the  fact  that, 
if  one  transformer  is  short-circuited,  the  voltage  across  the  other 
two  increases  by  73  per  cent,  unless  the  primary  and  generator 
neutrals  are  grounded.  In  any  case  if  one  transformer  of  the 
group  is  damaged  the  whole  group  must  be  shut  down. 

320.  Delta-star  and  Star-delta  Connections. — The  delta-star 
connection  is  used  at  the  generating  end  of  high-voltage  trans- 
mission lines  and  the  star-delta  connection  at  the  receiving  end. 

The  advantage  of  the  star  connection  for  the  high-voltage  side 
is  that  very  high  line  voltages  can  be  obtained  with  moderate 
transformer  voltages,  the  ratio  of  the  line  voltage  to  the  trans- 
former voltage  being  \/3:l,  the  cost  of  insulating  the  trans- 
former is  therefore  less  than  if  it  were  designed  for  the  full-line 
voltage  and  the  conductors  being  designed  for  larger  currents  are 
of  greater  section  and  better  able  to  stand  mechanical  strains. 

In  stepping  up  with  the  delta-star  connection  the  required 
third  harmonic  exciting  current  can  flow  in  the  three  transformers 
and  the  induced  voltages  will  be  sine  waves,  if  the  impressed  vol- 
tages are,  and  the  secondary  neutral  can  be  grounded  without 
causing  any  inductive  disturbances. 

When  stepping  down  with  the  star-delta  connection  the  re- 
quired third  harmonic  currents  cannot  flow  in  the  primary  but  the 
third  harmonic  e.m.fs.  induced  in  the  three  transformers  are  in 


TRANSFORMERS  391 

phase  and  cause  a  current  to  circulate  through  the  delta  and  this 
acts  as  a  magnetizing  current  and  supplies  the  missing  component 
of  flux.  The  exciting  current  is  therefore  divided  between  the 
primary  and  the  secondary. 

321.  Delta-delta  Connection. — This  connection  is  used  very 
extensively  even  in  the  case  of  very  high-voltage  transmission 
lines,  where  it  is  thought  advisable  to  make  provision  for  an  in- 
crease of  the  transmission  voltage  to  meet  the  requirements  of 
future  extensions  of  the  system.  In  such  a  case  the  secondary 
may  be  changed  to  star  and  extra  insulation  added  to  the  line. 

There  is  no  trouble  from  wave  distortion  as  in  the  star-star 
connection  and  if  one  transformer  burns  out  the  system  can  still 
be  operated  without  any  change  in  connections  with  the  two  re- 
maining transformers  in  open  delta  but  the  load  must  be  de- 
creased to  about  60  per  cent,  for  the  same  temperature  rise. 


a  &  A  B  aa&AU 

-/TrcwrosTON?    -/WOTOTN  'TOwsw^0      -/TOWWBN 

LJLJ       /    ™?  Uek*J     IT    IT 

assimjMiiLa  /  /  p^Msp^  fww^  £ 


C  C 

FIG.  366. — Open-delta   connection.  FIG.  367. — Tee   connection. 

322.  Open-delta  Connection. — If  one  transformer  of  a  delta- 
delta  bank  is  put  out  of  commission  the  system  can  still  be 
operated  with  the  other  two  transformers  without  any  change  in 
connections.  This  arrangement  of  two  transformers  on  a  three- 
phase  system,  Fig.  366,  is  called  the  "open-delta."  connection. 
It  may  be  used  as  an  emergency  connection  in  case  of  accident 
or  it  may  be  employed  in  a  new  installation  for  temporary  opera- 
tion and  a  third  transformer  added  to  complete  the  delta-delta 
bank  when  conditions  warrant  the  increase  in  capacity. 

In  this  system  each  transformer  carries  the  full-line  current 
and  for  a  given  output  the  transformer  currents  in  the  open-delta 
bank  will  be  73  per  cent,  greater  than  in  the  closed- delta  bank. 
To  supply  a  load  of  300  kva.  from  a  three-phase  system,  three 
100-kva.  transformers  are  required  connected  delta-delta  or 
two  173-kva.  transformers  connected  in  open-delta  and  therefore 
the  open-delta  bank  must  have  a  capacity  of  2  X  173  =  346  kva., 


392 


ELECTRICAL  ENGINEERING 


which  is  15  per  cent,  greater  than  the  closed-delta  bank.     Thus 
when  one  transformer  of  a  delta-delta  bank  burns  out  the  capacity 

100 
of  the  bank  is  reduced  to  -z-^  =  5%  Per  cent. 

Fig.  368  shows  a  vector  diagram  for  an  open-delta  system  with 
a  balanced  inductive  load.  E\,  E2  and  Es  are  the  e.m.fs.  gen- 
erated in  the  three  transformers  of  a  closed  delta;  they  are  also 
the  terminal  e.m.fs.  at  no  load  in  the  open  delta.  7'i,  7'2  and  7'3 
are  the  load  currents  in  the  closed  delta. 


FIG.  368. — Vector  diagram  for  an  open  delta. 

Transformer  No.  3  is  disconnected  and  the  two  remaining 
transformers  carry  currents  /i  =  I\  —  7'3  and  ^2=  I'z  —  /Y 
The  terminal  e.m.f.  of  No.  1  is  EAB  =  EI  —  IiZ,  where  Z  is  the 
impedance  of  each  of  the  transformers:  the  terminal  e.m.f.  of 
No.  2  is  EEC  =  EZ  —  I2Z  and  the  e.m.f.  between  the  lines  C  and 
A  is  EGA  =  —  (EAB  +  EBC)  since  the  sum  of  tKe  three  must 
equal  zero. 

The  three  terminal  e.m.fs.  are  slightly  unbalanced  under  load 
but  not  enough  to  have  any  serious  effect  on  operation.  If  an 
extra  impedance  =  %Z  is  added  to  the  primary  at  b  the  three 
secondary  line  e.m.fs.  will  be  equal.  Here  Z  is  the  impedance  of 
one  transformer. 

The  impedance  drops  in  Fig.  368  are  very  much  exaggerated 
and  the  terminal  e.m.fs.,  therefore,  appear  to  be  very  badly 
unbalanced. 

•  323.  "T"  Connection.— The  "T"  connection  is  a  second 
method  of  obtaining  a  three-phase  to  three-phase  transformation 
using  only  two  transformers.  One  transformer  is  designed  for 
only  87  per  cent,  of  the  three-phase  e.m.fs.  while  the  second  one 


TRANSFORMERS 


393 


is  designed  for  100  per  cent,  and  has  a  tap  at  its  center  point 
to  which  one  terminal  of  the  87  per  cent,  winding  is  connected 
(Fig.  367).  Between  the  lines  A,  B,  C  three  equal  e.m.fs.  are 
obtained  at  no  load  and  they  become  only  slightly  unbalanced 
under  load.  If  the  short-circuit  impedance  of  the  87  per  cent, 
transformer  is  equal  to  one-half  the  impedance  of  the  100  per 
cent,  transformer  the  e.m.fs.  will  be  balanced.  External  imped- 
ance may  be  added  to  obtain  this  condition. 

The  two  halves  of  the  winding  of  the  100  per  cent,  transformer 
carry  currents  which  are  out  of  phase  and  they  should  be  well 
interleaved. 

324.  Transformation  from  Two-phase  to  Three-phase. — The 
Scott  connection,  Fig.  369,  is  an  arrangement  of  two  transformers 
by  means  of  which  three-phase  power  can  be  obtained  from  a 
two-phase  circuit.  This  connection  is  reversible. 


EOAI3    E, 


EFB 


FIG.  369. — Two-phase  to  three-phase  (Scott  connection). 

Two  similar  transformers  are  used  with  their  primaries  con- 
nected to  the  two-phase  system.  The  secondary  AB  of  one  has 
a  tap  at  its  center  point  and  the  secondary  DC  of  the  other  has  a 
tap  86.6  per  cent,  from  the  end  C.  The  two  taps  are  connected 
at  F  and  three-phase  power  can  be  taken  off  from  the  terminals 
A,  B  and  C.  The  section  DF  is  not  used. 

ff  the  e.m.f.  generated  in  the  winding  DC  is 

eDc  =  Em  sin  d, 
and  that  in  A  B  is 

eAB  =  Em  sin  (0  -  90), 

then  the  e.m.fs.  generated  in  the  various  sections  are 
epc  =  —2~Em$me, 
eFA  =  M  Em  sin  (0  +  90) 

f>  1  /       Jj1  ,-ii-r-v       fQ  (\f\\    • 

"FB  —  /2  E'm  siu.  \v  —  \y\jj  f 


394  ELECTRICAL  ENGINEERING 

100 
the  neutral  point  of  the  three-phase  system  is  at  0,  which  is 

57.7  per  cent,  from  the  end  C,  and  thus 

Tji 

e0c  =  —  -p-  sin  0 
"V  3^ 

eFO  =  (  ^  ---  -^  }Em  sin  6  =  -  -=  Em  sin  0. 
\  2          -v/3/  2V3 

The  three-phase  star  voltages  are  eoA,  eOB  and  eoc', 
eOA  =  eQI,  +  ePA  =  -  ~~  sin  0  +        sin  (0  +  90) 


-^W  —  -3  cos  0  -  ^  sin  0    =  -p  sin  (0  +  120) 
\/3\  2 


eFB  =  -  --^=,  sin  0  +  -^  sin  (0  -  90) 
2\/3  2 


=      |(- 


cos  0  -  J$  sin  •-  -        sin  (0  +  240). 


Thus  the  three  star  voltages  COA,  COB  and  eoc  are  equal  in  value 
and  displaced  in  phase  by  120  degrees  and  form  a  three-phase 
system. 

These  star  voltages  may  be  combined  to  form  the  three  delta 
or  line  voltages,  eAB  =  Em  sin  (0  —  90),  eBc  =  Em  sin  (0  +  30) 
and  eCA  =  Em  sin  (0  +  150). 

If  the  loads  on  the  three  phases  are  balanced  the  terminal  vol- 
tages will  be  very  slightly  unbalanced,  since  the  impedance  from 
0  to  the  two  terminals  A  and  B  is  greater  than  that  from  0  to  C. 
To  correct  this  an  external  impedance  Zc  may  be  added  at  C  and 
its  value  must  be  such  that 

Zc  +  ZQC  —  ZOF  ~\~ZFA 
and  taking  Z  as  the  short-circuit  impedance  of  each  transformer 


Z    _/V3          1 

vf~\2~  •  v§ 


z  -ff 


2        2V3       2 
and 

Zc  =  -^  +-  --  ^L  =  -       ~  l  Z  =  21.2  per  cent,  of  Z. 

2V3       2       V3          2  V3 

Fig.  369  shows  a  vector  diagram  for  a  balanced  load  of  power 
factor  cos  0. 

The  power  delivered  by  the  secondary  DC  is 


Pi  =  Ere  1  1  cos  0  =      -  El  cos  0 


•3  -  Co         D      -^^- 

4 


TRANSFORMERS  395 

where  E  and  I  are  the  line  voltage  and  line  current  respectively. 
The  total  power  is  \/3EI  cos  6  and  thus  each  transformer  supplies 
one-half  of  the  load. 

ULj6 

f 

3^       S4 

Diametrical  ox  Double-Star  Ring 

FIG.  370. — Three-phase  to  six-phase. 

325.  Single-phase  Power  from  Three-phase  Circuits. — Single- 
phase  power  can  be  drawn  from  any  phase  of  a  three-phase  circuit 
but  when  the  amounts  of  power  required  are  large  this  leads  to  an 
unbalance  of  the  terminal  voltages  and  to  unequal  heating  of  the 
three  phases.  Fig.  371  shows  one  method  of  connecting  three 
transformers  on  a  three-phase  four-wire  system  to  supply  a  single- 
phase  load  by  means  of  which  equal  currents  are  drawn  from  the 
three  phases.  The  power  factors  of  the  loads  on  the  three  trans- 
formers are  very  different  but  the  temperature  rise  will  be  the 
same.  The  single-phase  voltage  is  only  double  the  voltage  of  one 
transformer  and  therefore  the  transformer  capacity  must  be  150 
per  cent,  of  the  load. 

Thtee-Pkase,  Four  Wire 

A — T 


njoTOn  njwftn  rowoci 
no  LJ  110  I  1  110  I 

1        Bitigle-Ehase        * 

FIG.  371. — Three-phase   to   single-phase. 

326.  Multiple  Operation  of  Transformers. — In  order  that  two 
transformers  connected  in  multiple  shall  divide  the  load  in  pro- 
portion to  their  capacities  it  is  necessary  (1)  that  they  have  the 
same  ratio  of  turns  and  (2)  that  the  per  cent,  resistance  drops 
and  the  per  cent,  reactance  drops  are  the  same  in  both. 

If  the  ratios  are  not  the  same  they  may  be  equalized  by  using 
an  auto-transformer  and  differences  in  the  impedances  may  be 
corrected  by  adding  external  impedance  to  one  transformer. 

In  three-phase  systems  only  those  groups  of  transformers  can 


396  ELECTRICAL  ENGINEERING 

be  operated  in  multiple  which  give  the  same  change  in  phase  be- 
tween the  primaries  and  secondaries,  a  star-star  bank  can  be 
operated  in  multiple  with  a  similar  star-star  bank  or  with  a 
delta-delta  bank,  but  neither  of  these  banks  can  be  operated  in 
multiple  with  a  delta-star  or  star-delta  bank.  One  delta-star 
bank  may  be  operated  in  multiple  with  another  delta-star  bank, 
or  it  may  be  operated  in  multiple  with  a  star-delta  bank  with 
its  secondary  voltages  reversed,  if  the  ratios  are  such  as  to  make 
the  voltages  equal. 

327.  Booster  Transformers. — Booster  transformers  are  con- 
nected with  their  primaries  across  the  line  and  their  secondaries 

. in  series  with  the  line  to  raise  or 

lower  the  voltage  (Fig.  372). 

If  the  primary  becomes  discon- 
nected while  the  load  is  -jm,  the 
booster  transformer  becomes  *a  series- 

^  n_ 

FIG.  372.— Booster  transformer,     transformer    with    open   secondary. 

The  load  current  acts  as  an  exciting 

current  and  very  high  voltages  may  be  generated  in  the  high- 
voltage  winding,  resulting  in  danger  to  persons  and  damage  to 
the  insulation.  To  disconnect  a  booster  transformer  under  load 
the  secondary  must  be  short-circuited  at  the  same  instant  that 
the  primary  is  opened. 

328.  Auto -transformers. — An  auto-transformer  has  only  one 
winding.     The  primary  includes  all  the  turns,  while  the  second- 
ary includes  only  a  part  of  them.     The  secondary  voltage  is  usu- 
ally made  variable  by  bringing  out  a  number  of  taps. 

Auto-transformers  may  be  used  in  almost  every  case  in  which 
ordinary  transformers  are  used  but  are  not  satisfactory  where  it 
is  desirable  to  have  the  secondary  insulated  from  the  primary. 

They  are  used  very  extensively  to  obtain  a  variable  voltage 
for  starting  induction  motors,  synchronous  motors,  single-phase- 
series  motors,  etc.,  and  as  balance  coils  on  three-wire  distributing 
circuits.  They  are  necessary  when  transformers  of  unequal  ratios 
are  to  be  operated  in  multiple.  Three-phase  auto-transformers 
are  often  used  to  interlink  two  systems  of  unequal  voltage. 

Fig.  373  shows  an  auto-transformer  for  starting  and  operating 
single-phase  series  motors  from  a  high-voltage  trolley  in  electric- 
railway  service.  The  section  of  the  winding  between  T  and  M 
carries  the  primary  current,  which  is  small,  while  the  section 
from  M  to  R  carries  the  difference  between  the  secondary  and 


TRANSFORMERS 


397 


primary  currents  which  is  very  large  if  the  ratio  of  voltages  is 
large  and,  therefore,  it  must  be  made  of  large  current  capacity. 

The  copper  loss  in  an  auto-transformer  is  smaller  than  in  an 
equivalent  two-coil  transformer  and  the  efficiency  is  therefore 
higher  but  this  advantage  decreases  as  the  ratio  of  turns  increases. 

Auto-transformers  may  be  used  either  to  step  up  or  step  down 
the  voltage. 

When  starting  induction  motors  or  synchronous  motors,  a  low 
voltage  must  be  impressed  at  start  to  prevent  objectionable 
current  surges.  From  40  to  70  per  cent,  of  full  voltage  is  usually 
impressed  at  the  first  step  depending  on  the  size  of  motor. 
Starters  for  medium-sized  motors  will  usually  have  only  two  taps 
while  for  large-sized  motors  three  or  more  taps  may  be  required. 

T  Trolley  3 


R  Rail 

Auto-Transfoimer  for 
Single- Phase  Electric 
Eailway  Service 


Star- Connected 
Auto-Starter 


FIG.  373. 


FIG.  374. 


Auto -Transformer 
as  Balance  Coil  for 
a  Three- Wire  System 

FIG.  375. 


For  three-phase  motors  three  auto-transformers  connected  in 
star  may  be  used  or  two  connected  in  open  delta  or  tee  (Fig.  374). 

Fig.  375  shows  an  auto-transformer  used  as  a  balance  coil  on  a 
220-volt,  three-wire  system. 

329.  Instrument  Transformers. — Instrument  transformers  are 
of  two  kinds,  shunt  or  potential  transformers  and  series  or  current 
transformers. 

Potential  transformers  are  used  to  reduce  the  line  voltage  to  a 
value  suitable  for  operating,  indicating,  integrating  and  record- 
ing instruments  as  voltmeters,  wattmeters,  power-factor  meters, 
etc.,  and  relays  for  voltage  regulation  or  protection.  They  also 
serve  to  insulate  these  instruments  from  the  high-voltage  circuit. 

The  ratio  of  primary  impressed  voltage  to  secondary  terminal 
voltage  should  be  as  nearly  as  possible  constant  and  the  phase 
angle  between  these  two  voltages  should  be  small.  Fig.  376 
shows  a  vector  diagram  for  a  potential  transformer  with  a  1  to  1 
ratio  and  non-inductive  load.  This  is  the  usual  case  since  the 
voltage  coils  of  instruments  are  practically  non-inductive.  The 
same  symbols  are  used  as  in  Art.  302. 


398 


ELECTRICAL  ENGINEERING 


E'  . 


The  ratio  ^r  is  constant  and  is  equal  to  the  ratio  of  turns  but 
Ez 

f  JT' 

the  ratio  ^  is  always  greater  than  -g-  on  account  of  the  imped- 
ance drops  in  the  primary  and  secondary.  Thus  to  obtain  the 
primary  impressed  voltage  E\  from  the  secondary  terminal  volt- 
age E}  it  is  necessary  to  multiplyEby  the  nominal  ratio  and  by  a 


FIG.  376. — Instrument  transformer. 


FIG.  377. 


ratio  correction  factor  ki.  This  factor  varies  with  the  impressed 
primary  voltage  or  secondary  terminal  voltage  and  it  also  varies 
with  the  secondary  load,  that  is,  with  the  number  and  type  of 
instruments  connected  to  the  secondary.  The  ratio  correction 
factor  is  usually  plotted  on  a  base  of  secondary  terminal  voltage 
for  a  specified  secondary  load  resistance.  Fig.  378  shows  values 
of  ki  for  a  Weston  portable  transformer  it  increases  with  the 
secondary  terminal  voltage  E. 


50  75  100 

Secondary  Volts 

FIG.  378. — Ratio-correction  curves  for  a  Weston  portable  potential  transformer. 

The  angle  of  phase  shift  or  the  phase  angle  of  the  transformer 
under  given  conditions  is  the  angle  by  which  EI  and  E  are  out  of 
line  and  is  shown  as  7  in  the  diagram.  The  phase  angle  depends 
on  the  impedance  drops  in  the  transformer  and  increases  with  the 
secondary  load  current  and  with  the  secondary  terminal  voltage. 
The  phase  angles  of  standard  potential  transformers  are  very 
small,  unless  they  are  overloaded,  and  they  can  be  neglected 
except  in  the  most  accurate  measurements  or  at  low  power  fac- 


TRANSFORMERS  399 

tors.  Phase-angle  corrections  do  not  apply  to  voltmeter  read- 
ings but  only  to  the  indications  of  wattmeters  and  power-factor 
meters  which  are  affected  by  change  of  phase. 

Potential  transformers  are  designed  with'  very  low  impedances 
by  subdividing  the  coils  well  and  placing  them  as  close  together 
as  possible.  The  very  best  insulating  materials  must  be  used. 
They  are  designed  with  very  low  flux  densities  to  prevent  distor- 
tion of  wave  form  due  to  the  exciting  current. 

Calibrations  must  be  made  with  a  definite  instrument  load  in 
the  secondary. 

330.  Current  Transformers. — Current  transformers  are  con- 
nected with  their  primaries  in  series  with  the  circuit  and  are  used 
to  reduce  the  line  current  to  a  value  suitable  for  current  relays 
and  for  indicating  and  integrating  instruments.  They  also 
insulate  the  instruments  from  the  high-voltage  circuits. 

The  ratio  of  primary  to  secondary  current  should  be  as  nearly 
as  possible  constant  and  the  phase  angle  between  them  should 
be  very  small. 

Fig.  376  may  also  be  used  as  a  vector  diagram  for  a  current 
transformer.  The  meter  load  is  again  assumed  to  be  non-induc- 
tive and  its  resistance  is  given  as  the  ratio  of  secondary  terminal 

ET 

voltage   to   secondary   current  =  j-.     The  exciting  current  /o 

*2 

added  to  the  primary  load  current  prevents  the  ratio  of  currents 
from  being  constant  and  equal  to  the  ratio  of  turns  and  it  also 
causes  the  currents  to  be  out  of  line  by  the  phase  angle  /?. 

The  exciting  current  can  be  kept  low  by  using  iron  of  very 
high  permeability  and  low  loss  and  designing  the  secondary  with 
a  low  impedance.  The  core-loss  current  is  nearly  in  phase  with 
the  primary  load  current  and  has  a  large  effect  on  the  ratio  but 
little  effect  on  the  phase  angle  while  the  magnetizing  current  has 
a  large  effect  on  the  phase  angle  and  a  small  effect  on  the  ratio. 
Increasing  the  secondary  load  resistance  by  adding  extra  instru- 
ments increases  the  voltage  generated  in  the  secondary  for  a  given 
value  of  secondary  current  and  so  increases  the  flux  required 
and  the  exciting  current.  This  increases  both  the  phase  angle 
and  the  ratio  correction.  Reactance  added  to  the  secondary  cir- 
cuit causes  the  exciting  current  to  swing  more  nearly  into  line 
with  the  load  current  and  therefore  increases  the  ratio  correction 
but  decreases  the  phase  angle. 

Fig.  379  shows  ratio  and  phase  angle  curves  for  a  Westo.n 


400 


ELECTRICAL  ENGINEERING 


standard  current  transformer.  To  find  the  primary  current, 
corresponding  to  a  given  secondary  current  I 2,  multiply  7  2  by  the 
nominal  ratio  and  by  the  ratio  correction  factor  &2. 

The  ratio  correction  factors  and  phase  angles  are  smaller  on 
60  cycles  than  on  25  cycles  since  the  flux  and  exciting  currents 
for  a  given  load  current  are  less.  The  phase  angles  of  current 
transformers  are  greater  than  those  of  potential  transformers 
and  a  correction  should  always  be  made  for  them. 


01  2345 

Secondary  Amperes 

FIG.  379. — Ratio-correction   and   phase-angle   curves   for    a   Weston    portable 
current  transformer. 

The  primary  winding  of  a  current  transformer  usually  consists 
of  a  single  turn  in  series  with  the  line  while  the  secondary  may 
have  any  number  of  turns  depending  on  the  reduction  in  current 
required.  The  current  in  the  primary  is  the  load  current  of  the 
circuit  and  does  not  depend  in  any  way  on  the  transformer.  It 
sets  up  a  magnetic  flux  in  the  core  which  links  with  both  the  pri- 
mary and  secondary  windings.  If  the  secondary  were  open- 
circuited  all  the  primary  m.m.f .  would  be  magnetizing  and  would 
produce  a  very  large  flux  in  the  core  and  in  a  transformer  having 
a  large  ratio  thousands  of  volts  would  be  induced  in  the  secondary, 
resulting  in  danger  to  operators  and  damage  to  the  instrument. 
The  iron  core  is  designed  to  carry  only  the  flux  required  with 
closed  secondary  and  it  would  be  very  highly  saturated  under 
open-circuit  conditions  and  would  become  hot  due  to  excessive 
iron  losses.  The  transformer  would  require  to  be  recalibrated 
before  its  readings  could  be  relied  on.  Current  transformers 
should  always  be  provided  with  a  short-circuiting  link  which 
must  be  closed  before  the  secondary  is  opened. 


TRANSFORMERS  401 

Potential  transformers  are  built  for  all  voltages  up  to  100,- 
000  volts  but  for  the  high  voltages  they  become  very  expensive. 
The  secondary  is  usually  designed  for  100  volts.  • 

Current  transformers  are  built  with  a  rating  of  about  25  volt- 
amp,  and  a  secondary  current  of  5  amp.  The  primary  current 
ranges  from  5  amp.  to  3,000  or  more.  On  circuits  up  to  20,000 
volts  they  are  usually  air-cooled  but  above  this  they  are  oil- 
immersed.  For  primary  currents  above  600  amp.  no  primary 
winding  is  supplied  but  the  transformer  is  slipped  over  the 
bus-bar  which  then  forms  its  single  turn. 

331.  Example. — In  the  circuit  in  Fig.  377(o)  it  is  required  to  find  the 
primary  voltage  E\,  current  I\,  power  factor  cos  61  and  power  Wi,  knowing 
the  secondary  voltage  E,  current  /  and  power  W.     The  phase  angle  of  the 
potential  transformer  is  7,  Fig.  376,  and  the  ratio  correction  factor  is  fcij 
the  phase  angle  of  the  current  transformer  is  /3  and  the  ratio  correction 
factor  is  k%. 

The  primary  voltage  is  E\  —  E  X  nominal  ratio  X  k\. 
The  primary  current  is  /i  =  /  X  nominal  ratio  X  kz. 

W  W 

The  secondary  power  factor  is  cos  6  =  ~^j,  or  6  =  cos  l  -prf 

Referring  to  Fig.  377(6)  the  primary  angle  of  lag  is  d\  =  9  +  0  +  7  and  the 
primary  power  factor  is  cos  (6  +  /3  +7). 

The  secondary  voltage  always  lags  behind  the  primary  voltage  and  the 
secondary  current  always  leads  the  primary  current. 

Errors  due  to  the  phase  angles  /3  and  7  are  much  more  serious  on  low  power 
factors  than  on  high  power  factors.  When  using  two  wattmeters  to  read 
three-phase  power,  phase-angle  corrections  should  always  be  made. 

332.  The  Constant-current  Transformer. — The  constant-cur- 
rent transformer  is  shown  diagrammatically  in  Fig.  380.     The 
primary  coil  P  is  fixed  in  position  and  receives  power  at  constant 
voltage.     The  secondary  coil  S  is  movable  and  regulates  for 
constant  current  in  the  receiver  circuit  which  it  supplies  irrespec- 
tive of  the  load.     The  transformer  is  used  to  obtain  a  constant 
current  for  series  arc-light  circuits. 

When  the  secondary  coil  is  close  to  the  primary  there  is  very 
little  leakage  and  most  of  the  flux  produced  by  the  primary  links 
with  the  secondary  and  the  secondary  voltage  is,  therefore,  a 
maximum.  Primary  and  secondary  currents  are  in  opposite 
directions  and  the  two  coils  repel  one  another.  The  weight  W 
is  so  adjusted  that  the  pull  due  to  it  together  with  the  force  of 
repulsion  of  the  coils  just  balances  the  weight  of  S  and  allows 
the  coil  to  remain  in  such  a  position  that  the  required  current 
flows  in  it.  Fig.  381  shows  the  flux  in  the  core. 

26 


402 


ELECTRICAL  ENGINEERING 


If  the  resistance  or  impedance  of  the  load  circuit  decreases  due 
to  the  cutting  out  of  one  or  more  arc  lamps  an  increase  of  the 
current  in  both  secondary  and  primary  follows  and  the  repulsion 
between  the  coils,  which  is  proportional  to  the  product  of  their 
currents,  increases.  The  secondary,  therefore,  rises  and  increases 
the  leakage  reactances  of  both  coils  and  so  less  of  the  primary 
magnetism  links  with  the  secondary;  its  voltage  is,  therefore, 
decreased  and  its  current  drops  to  the  required  value.  The 
moving  arm  must  be  designed  to  give  the  required  regulation  with 
a  fixed  weight  W. 


FIG.  380. — Constant-current  trans- 
former. 


FIG.  381. — Flux  in  the  core  of  a  con- 
stant-current   transformer. 


Such  an  arrangement  regulates  for  constant  current  between 
the  limits  of  secondary  voltage  set  by  the  two  extreme  positions 
of  the  moving  coil. 

Neglecting  the  primary  exciting  current  the  relation  between 
the  terminal  voltages  may  be  written  : 


Eij  /2,  7*1  and  r2  are  constant  and  Xi  and  xz  increase  as  E  decreases. 

333.  Regulation  and  Power  Factor.  —  The  secondary  current 
should  be  maintained  within  0.1  amp.  of  the  rated  current  from 
no  load  to  full  load. 

The  full-load  power  factor  of  an  alternating  arc-lamp  system 
seldom  exceeds  70  per  cent,  and  is  much  lower  at  light  loads. 
In  series  tungsten-lamp  systems  the  power  factor  varies  from 
about  23  per  cent,  at  one-fourth  load  to  85  per  cent,  at  full  load. 


TRANSFORMERS 


403 


334.  Cooling. — Small  constant-current  transformers  are  cooled 
by  natural  air  circulation;  above  100  kva.  they  are  oil-immersed 
and  may  be  water-cooled. 

335.  Induction  Regulator. — Induction  regulators  are  special 
transformers  used  to  vary  the  voltage  of  an  alternating-current 
distributing  circuit  or  the  voltage  impressed  on  a  rotary  con- 
verter. 

/Secondary  Coll  S 

^Primary  Coil  P 


FIG.  382.  FIG.  383. 

Single-phase  induction  regulator. 


FIG.  384. 


There  are  two  types  of  induction  regulators,  single-phase  and 
polyphase. 

1.  The  single-phase  regulator  is  illustrated  in  Figs.  382  to  384. 
The  primary  coil  P  is  carried  on  a  movable  core  built  of  laminated 
iron  and  is  connected  across  the  line.  The  secondary  coil  S  is 
carried  on  a  stationary  core  and  is  connected  in  series  with  the 
line  to  raise  or  lower  the  voltage  (Fig.  385). 

Secondary 


Line 


Load 


FIG.  385. 


S2  Si 
FIG.  386. 


The  primary  exciting  current  produces  an  alternating  flux, 
which  induces  a  voltage  in  the  secondary.  This  secondary  volt- 
age varies  with  the  position  of  the  primary  winding,  but  it  is 
always  in  phase  with  or  in  opposition  to  the  impressed  voltage  or 
line  voltage.  In  Fig.  382  all  the  primary  flux  (neglecting  leakage) 
passes  through  the  secondary  coil  and  the  secondary  voltage  is  a 
maximum  and  is  in  opposition  to  the  impressed  voltage,  and  so 
gives  the  minimum  load  voltage. 

In  Fig.  386  OP  is  the  impressed  voltage  or  line  voltage,  OSi  is 


404  ELECTRICAL  ENGINEERING 

the  maximum  secondary  induced  voltage  and  S\P  is  the  load 
voltage. 

The  load  current  of  the  circuit  flows  in  the  secondary,  and 
there  must  also  be  a  load  current  in  the  primary  of  equal  and 
opposite  m.m.f.  as  in  the  ordinary  transformer. 

When  the  movable  core  is  turned  through  an  angle  6,  Fig.  383, 
only  part  of  the  primary  flux  passes  through  the  secondary  and 
the  secondary  voltage  is  reduced  approximately  in  the  ratio 
1  :  cos  0,  but  is  still  in  opposition  to  the  line  voltage.  The  load 
voltage  is  represented  by  S^P,  Fig.  386. 

With  the  primary  coil  at  right  angles  to  the  secondary  coil, 
Fig.  384,  none  of  the  primary  flux  passes  through  S  and  there  is 
no  secondary  induced  voltage. 

When  the  core  is  turned  through  180  degrees  the  secondary 
voltage  is  again  maximum,  but  is  in  phase  with  the  impressed 
voltage  and  so  raises  the  load  voltage  to  S$P.  Thus  the  total 
variation  of  the  load  voltage  is  from  SiP  to  S&P  and  is  equal  to 
twice  the  maximum  secondary  voltage. 

In  Fig.  384  the  m.m.f.  of  the  load  current  in  the  secondary 
cannot  be  opposed  by  any  primary  m.m.f.  since  the  coils  are  at 
right  angles.  To  supply  the  m.m.f.  required  to  balance  the 
secondary  load  m.m.f.,  and  so  prevent  a  large  reactance  drop  in 
the  winding,  the  coil  T  called  the  "tertiary"  coil  is  placed  on  the 
movable  core  at  right  angles  to  the  primary  coil.  It  is  short- 
circuited  and  exerts  a  m.m.f.  equal  and  opposite  to  the  second- 
ary m.m.f.,  and  so  reduces  the  secondary  reactance  to  the  value 
corresponding  to  the  leakage  flux. 

The  only  current  carried  by  the  primary  coil  in  this  position 
is  the  exciting  current. 

In  intermediate  positions  of  the  rotor  the  secondary  m.m.f.  is 
partly  balanced  by  the  induced  current  in  the  tertiary  or  com- 
pensating coil  and  partly  by  a  load  current  in  the  primary  coil. 

2.  The  polyphase  induction  regulator  has  a  polyphase  wind- 
ing on  the  moving  core,  which  is  connected  to  the  polyphase 
supply.  The  secondary  or  stator  is  wound  with  the  same  number 
of  phases,  but  the  phase  windings  are  kept  entirely  separate  so 
that  they  can  be  connected  in  the  different  lines  to  raise  or  lower 
the  voltage. 

When  polyphase  currents  flow  in  the  primary  windings  a 
revolving  magnetic  field  is  produced  of  constant  value  as  in  the 
alternator  or  induction  motor.  This  field  cuts  the  secondary 


TRANSFORMERS  405 

windings  and  generates  e.m.fs.  in  them  of  the  same  frequency  as 
the  primary  impressed  e.m.fs.,  but  less  in  the  ratio  of  turns. 

As  the  primary  is  turned  the  magnitude  of  the  revolving  field 
is  not  changed,  and  therefore  the  magnitudes  of  the  secondary 
e.m.fs.  are  not  changed  but  their  phase  relations  with  the  im- 
pressed e.m.fs.  are  changed  and  the  load  voltage  is  varied  as 
shown  in  Fig.  387.  By  turning  the  primary  through  180  degrees 
the  phase  of  the  secondary  e.m.f .  is  changed  from  direct  opposition 
to  the  line  voltage  to  direct  addition  to  it  and  thus  the  load  volt 
tage  is  varied  by  an  amount  equal  to  twice  the  secondary  voltage. 

The  rotor  must  be  clamped  in  the  required  position  or  it  will 
tend  to  rotate  at  a  high  speed  in  the  direction  opposite  to  that  of 
the  revolving  field. 


0  Si 

FIG.  387. 

The  advantage  of  the  induction  regulator  over  a  transformer 
with  variable  voltage  taps  on  the  secondary  or  primary  is  that 
the  variation  of  voltage  is  uniform  over  the  entire  range.  Regu- 
lators are,  however,  very  expensive  and  require  a  large  exciting 
current  and  have  large  leakage  reactances. 

The  regulator  is  operated  either  by  hand  Or  by  means  of  a 
small  motor  placed  on  the  top  or  it  can  be  made  automatic  if 
required. 

336.  Design  of  a  Transformer.  —  Given  the  type  and  method 
of  cooling,  number  of  phases,  frequency,  capacity  in  kilovolt- 
amperes,  ratio  of  voltages  and  the  class  of  service,  a  transformer 
may  be  designed  using  the  equations  and  constants  listed  below. 

The  e.m.f.  equation  is 

E  =  4.44/n$10-8  volts          Art.  301.  (353) 


Volts  per  turn  =  e  =       g^put  in.  volt-amperes, 

A/ 

where  k  is  a  constant  for  which  the  following  approximate  values 
may  be  used  : 

For  shell-type  power  transformers,  k  =  25, 
For  core-type  power  transformers,  k  =  50, 
For  core-type  distributing  transformers,  k  =  80. 


406 


ELECTRICAL  ENGINEERING 


Flux  densities.  The  flux  densities  to  be  used  depend  on  the 
quality  of  the  iron,  the  frequency,  and  the  service  in  which  the 
transformer  is  to  be  placed. 

With  good  alloyed  iron  densities  from  65,000  to  90,000  are 
used.  The  lower  values  should  be  used  for  60-cycle  distributing 
transformers  to  keep  down  the  iron  losses.  When  using  silicon 
steel  with  its  low  iron  losses  higher  densities  may  be  employed 
but  care  must  be  taken  that  the  magnetizing  current  does  not 
become  too  large  since  the  permeability  of  silicon  steel  is  compara- 
tively low. 

Ratio  of  losses.  In  distributing  transformers  the  core  loss  is 
from  30  to  50  per  cent,  of  the  copper  loss  while  in  power  trans- 
formers the  two  are  approximately  equal. 

Knowing  the  voltage  and  the  volts  per  turn  the  number  of 
turns  may  be  found  and  substituting  in  equation  (353)  the  flux 
3>  is  determined.  Dividing  3>  by  the  assumed  flux  density  gives 
the  section  of  the  magnetic  circuit.  Fig.  388  shows  the  magnetic 
circuits  of  shell-  and  core-type  transformers  and  their  relative 
dimensions. 


Watt*  par  Pound  Wait,  pal  Pound 

0     0.4  0.8  1.2  1.8  1.0  2.4  2.8  3.2  0    0.4  0.8  1.2  1.6  2.0  2.1  2.8  3.2  3.8 

100000 


4      8     12    IB    20    24    23    32    36    40 
Volt  Amp.ro  |  or  Pound 


Core  Type 


FIG.  388. 


FIG.  389. — Magnetic  characteristics  of 
transformer  sheets. 


Before  deciding  on  the  size  of  the  opening  in  the  iron  the  sec- 
tions and  arrangement  of  the  winding  must  be  determined. 
Design  of  the  windings. 

(a)  Volts  per  coil  should  not  exceed  5,000. 

(b)  Volts  per  layer  should  not  exceed  350. 

(c)  Current  density  should  lie  between  625  and  1,250  amp.  per 
square  inch  or  2,000  to  1,000  circ.  mils  per  ampere.     In  large 
water-cooled  transformers  higher  densities  may  be  used  up  to 
2,000  amp.  per  square  inch. 

(d)  The  strip  used  in  the  windings  should  not  be  wider  than 
0.5  in.  for  60  cycles  nor  0.75  in.  for  25  cycles  to  prevent  excessive 
eddy  currents  in  the  copper. 


TRANSFORMERS  407 

(e)  Coils  should  not  be  made  more  than  1  in.  in  thickness  and 
must  have  a  surface  in  contact  with  the  oil  of  at  least  2  to  3  sq. 
in.  per  watt  copper  loss,  the  lower  value  being  for  water-cooled 
transformers. 

Copper  losses  should  be  calculated  at  75°C.  and  if  they  are 
found  to  be  too  large  either  the  current  density  must  be  decreased 
or  the  radiating  surface  increased  by  further  subdivision  of  the 
winding.  Decrease  of  current  density  will  improve  the  efficiency 
while  an  increase  in  the  number  of  coil  groups  will  decrease  the 
reactance  and  therefore  improve  the  regulation  but  will  not  affect 
the  efficiency. 

(/)  The  spacings  between  coils  and  between  coils  and  iron  de- 
pend on  the  voltage  and  approximate  values  may  be  obtained 
from  Fig.  393.  The  fullerboard  insulation  between  the  high- 
voltage  (H.  V.)  and  low-voltage  (L.  V.)  coils  may  be  assumed  to 
take  up  from  one-third  to  one-half  the  available  space.  Spacing 
blocks  are  used  to  strengthen  the  coils  and  to  keep  the  paths  for 
oil  circulation  open.  Low-voltage  coils  are  always  placed  next 
the  iron  and  the  windings  are  arranged  in  groups  of  high-  and 
low-voltage  coils  to  reduce  the  reactance. 

Having  determined  the  space  required  for  the  windings  and  the 
opening  in  the  iron  the  size  of  plates  may  be  found  and  the  total 
volume  and  weight  of  iron  calculated. 

Fig.  389  shows  characteristic  curves  for  good  alloyed  iron  at 
25  and  60  cycles  plotted  with  flux  density  against  watts  iron  loss 
per  pound  and  volt-amperes  per  pound.  Knowing  the  flux  dens- 
ity the  watts  iron  loss  per  pound  may  be  obtained  and  this  value 
multiplied  by  the  total  weight  of  iron  gives  the  total  iron  loss  or 
core  loss  in  the  transformer.  If  the  iron  loss  is  too  large  it  may 
be  decreased  by  using  a  lower  flux  density  or  by  making  the  core 
of  silicon  steel  with  lower  losses.  If  the  total  flux  in  the  trans- 
former is  decreased  the  number  of  turns  must  be  increased  and 
this  will  result  in  increased  copper  losses. 

The  iron  must  have  a  radiating  surface  in  contact  with  the  oil 
of  at  least  1  sq.  in.  per  watt  lost  in  the  core. 

Self-cooled  transformers  are  placed  in  corrugated  tanks  and 
are  immersed  in  oil.  From  5  to  8  sq.  in.  of  tank  surface  are 
required  for  each  watt  lost  depending  on  the  depth  of  the 
corrugations. 

In  water-cooled    transformers  with   cooling  coils  \Y±  in-  m 


408 


ELECTRICAL  ENGINEERING 


diameter  a  coil  surface  of  1  sq.  in.  per  watt  lost  is  required  and 
about  Y±  gal.  of  water  per  minute  per  kilowatt  lost. 

337.  Reactance. — The  reactance  of  the  windings  of  a  trans- 
former can  be  determined  very  accurately  since  the  leakage  paths 
are  of  simple  form. 

Fig.  390  shows  a  section  through  the  coils  of  a  shell-type  trans- 
former and  an  enlarged  section  of  one  group  of  coils,  w  =  width 
of  the  iron  opening,  «i  and  a2  are  the  widths  of  high-voltage  and 
low-voltage  coils  respectively,  and  di  and  d2  are  the  thicknesses 
of  these  coils.  The  distance  between  the  coils  s.  When  ai 
and  a2  are  only  slightly  smaller  than  w  the  leakage  flux  may  be 
assumed  to  cross  the  opening  from  one  side  to  the  other  but  in 
high-voltage  transformers  where  a\  is  much  less  than  w,  a  more 
approximate  expression  for  the  length  of  the  leakage  path  in  air 
may  be  found  by  inspection. 


Core 


Section  of  Winding 


High  Voltage  Ooil 


FIG.  390. 


Assuming  that  the  two  windings  are  divided  into  the  same 
number  of  coils  =  g,  let  t\  be  the  number  of  turns  in  one  primary 
coil  carrying  current  /i  and  t2  be  the  number  of  turns  in  one 
secondary  coil  carrying  current  /2,  then  t-Ji  =  falz.  The  leakage 
flux  about  any  coil  may  be  separated  into  two  parts:  (a)  that 
crossing  the  coil,  and  (6)  that  passing  through  between  the  high- 
and  low-voltage  coils.  One-half  of  the  flux  between  the  coils  is 
assumed  to  surround  each  of  them. 

At  a  distance  y  from  the  inside  of  coil  A  the  m.m.f.  acting  is 

ti  Ii  -T  and  this  produces  a  flux,  in  the  path  of  section  l\dy  and 
«i 

length  w,  of  a  value 


d<t>' 


w 
hdy 


TRANSFORMERS  409 

Where  li  is  the  length  of  the  mean  turn  of  the  coil;  this  flux 
surrounds  only  -4^  turns  and  is  equivalent  to  a  flux  d<j>  surround- 
ing ti  turns,  where 

,   .  y    __  brtj 


and  the  flux  equivalent  to  the  total  flux  crossing  the  coil  is 
*  =  f4  =  ^  rVdy  =  ^  *• 

J^O  ttX*l2      JO  W          3 

The  flux  in  the  path  between  the  coils  is 


w 

and  the  part  of  this  surrounding  coil  A  is 


2          w        2 
The  total  leakage  flux  surrounding  coil  A  is 

and  the  self-inductance  of  the  coil  is 

£A  =  ^^  =  -        -  /  - -1  -f-  —  j  absolute  units,   if  all  dimensions 

are  in  centimeters. 

Using  inch  units,  the  inductance  of  each  coil  is 

£A  =  -       -  (~  +  -)2.54  absolute  units, 
iv       \  o         2i' 

or 

=  3.2  X  10~8  -^-^  (V  +  ^)  henry. 
If      \  o          2tl 

If  the  g  coils  are  connected  in  series  the  inductance  of  the  winding 
is 


and  the  reactance  is 

X,  -  27T/L!  =  27T/  X  3.2  X  lO-8^-1        +       flf 


=  20.2  X  10-8/—  -  (i1  +        ohms,  (355) 

g    w  \  6 


410 


ELECTRICAL  ENGINEERING 


where  tti  =  tig  =  total  primary  turns.     The  reactance  is  in- 
versely proportional  to  the  number  of  coil  groups. 
The  reactance  of  the  secondary  winding  is 

20.2X10-*/71?2--.  (t2  +  4' 


X 


IV 


ohms. 


(356) 


The  reactance  of  a  core-type  transformer  may  be  worked  out  in  a 
similar  way. 


FIG.  394. 


FIG.  395. 


FIG.  396. 


FIG.  397. 


338.  Design  of  a  Shell-type,  Water-cooled  Transformer, 
Single-phase,  60  Cycles,  1,000  kva.,  22,000  to  2,200  Volts  for 
Power  Transmission. — 


Volts  per  turn  =  Voutput  in  volt-amperes  = 

k 


25 


40. 


Number  of  primary  turns 


22,000 
40 


550,  assume  540.     Make  the  pri- 


mary winding  of  six  coils  four  of  92  turns  and  two  outer  coils  of  86  turns, 
this  leaves  space  for  extra  insulation  on  the  end  turns  to  take  care  of  strains 
due  to  switching  or  lightning. 

Circular  mils  per  ampere  =  1,000. 

Current  density  =  1,270  amp.  per  square  inch. 

T>-  1,000,000        ._. 

Primary  current  =  =  45.4  amp. 


Section  of  conductor  =  45.4  X  1,000  =  45,400  circ.  mils  =  0.0368  sq.  in. 

Size  of  strip  =  0.5  X  0.08  bare  or  0.515  X  0.095  d.c.c. 

Actual  current  density  =  1,170  amp.  per  square  inch. 

Actual  circular  mils  per  ampere  =  1,085. 

Insulation  between  layers  =  2  ply  0.007  fullerboard. 

Thickness  of  coil  =  di  =  0.515  in. 

Width  of  coil  =  ai  =  0.095  X  92  +  2  X  0.007  X  91  =  10.014  =  10.5  in. 
co  allow  for  warping  (Fig.  398). 

The  end  coils  have  5  ply  0.007  fullerboard  between  turns  1  to  11  and  3  ply 
between  turns  11  to  31  making  the  coil  width  10.5  as  before. 


TRANSFORMERS 


411 


Number  of  secondary  turns 


540 

~r     =  54. 


Number  of  secondary  coils  =  6. 
Turns  per  coil  =  18,  connect  coils  in  series  multiple. 
Secondary  current  =  454  amp. 
Current  per  coil  =  227  amp. 
Amperes  per  square  inch  =  1,270  as  in  primary. 
Section  of  conductor  =  0.2  sq.  in.  =  0.4  in.  X  0.5  in. 
Use  four  strips  0.4  X  0.125  in  parallel  separated  by  1  ply  0.007  fuller- 
board  and  taped  with  half-lapped  tape. 

Insulation  between  layers  =  2  ply  0.007  fullerboard. 


High  Voltage 
Coil  Section 


Low  Voltage 
Coil  Section 


FIG.  398. 


in.  block  +  2  (H)  in.  fullerboard  +  %  in.  block 


Thickness  of  low  voltage  coil  =  d2  =  0.425  in. 

Width  of  low  voltage  coil  =  a2  =  18(4  X  0.125  +  3  X  0.007  +  0.025)  -f 
17  X  2  X  0.007  =  10.066. 

Take  the  width  =  10.5  in.  to  allow  for  warping  (Fig.  398). 
Spacing 

H.V.  to  L.V.   = 
1.0  in. 

H.V.  to  H.V.  =  %  in.  block -f-^  in.  fullerboard  +  %  in.  block  =  %  in. 

L.V.  to  L.V.  =  %  in.  block. 

L.V.  to  iron     =  Y±  in.  block  +H  in.  fullerboard  +  %  in.  block  =  %  in. 

H.V.  to  iron    =  1  in.  (Fig.  393). 

H.V.  to  iron  at  top  and  bottom  of  cores  =  Z)  =  4  in.  (Fig.  393). 
Arrangement  of  coils  ) 

Opening  in  iron  =  12.5  X  20  |  Fig.  394. 

Size  of  plates  5  X  25  and  5  X  17.5     j 
Flux  density  =  80,000  lines  per  square  inch,  assumed. 
Ei  22,OOOX108 


Flux  required  = 
Section  of  iron 


15.3  X106 
80,000 


4.44X60X540  = 
=  191   sq.   in.  =  (5+5)XAX0.9    (Fig.  391). 


412  ELECTRICAL  ENGINEERING 

191 
Height  of  tongue  =  h  =  1Q       Q  Q  =  21.2  in. 


Weight  of  iron  =  2     30  X  22.5  -20  X  12.5  -  -  X4 

X21.2X0.9X0.28=  5,538  Ib. 

where,  0.28  is  the  weight  of  1  cu.  in.  of  iron  in  pounds. 
Watts  lost  per  pound  =  0.70  (Fig.  389). 
Total  iron  loss  =  0.70  X  5,538  =  3,877  watts. 

3  877 
Core-loss  current  =  Ic  =  22~000  =  °-176  amP- 

Volt-amperes  per  pound  =  11.5  (Fig.  389). 
Volt-amperes  at  no  load  =  11.5  X  5,538  =  63,680. 

Exciting  current  =  22  QOQ  =  ^°  =  2'89  amp' 

2  89 
Per  cent,  exciting  current  =TK~!  X  100  =  6.36  per  cent,  of  full-load  current. 

Magnetizing  current  =  IM  =  Vl*  -/J  =  ^(2.89)2  -  (0.176)2  =2.88  amp. 

3  877 
No-load  power  factor  cos  00  =  ,>»  6QQ  100  per  cent.  =  6  per  cent. 

T  fy    OQ 

Primary  exciting  admittance  =  Y0  =  Vg2Q  -j-  &J  =  ^  =  22QQQ  =  1.31  X 

io-4. 

Conductance  =  g0  =  —•  =  22  OQQ  =  0.08  X  10~4. 
Susceptance  =  60  =  j=r  =  =  1.3  X  10~4. 


Radiating  surface  of  the  iron  =  8,900  sq.  in.  (Fig.  395). 

8  900 
Square  inches  per  watt  iron  loss  =     '  --  =  2.3. 

OjO  I   / 

Length  of  the  mean  turn  of  high-voltage  winding  =  2K  +  2L  +  irai 
=  117  A  in.  =  9.8  ft.  (Figs.  391  and  395). 

Resistance  of  the  H.V.  winding  at  75°C.  =  n  =  p  -.  -  rr-    =  13.2 

ClrC.  mils* 

9.8  X  540          1  ACt    , 
lM5~X45A  =  L42  Ohms' 

Primary  copper  loss  =  I^n  =  (45.4)  2  X  1.42  =  2,940  watts. 
Length  of  the  mean  turn  of  L.V.  winding  =  9.8  ft. 

Resistance  of  the  L.V.  winding  =  r2  =  T7J2  =  0.0142  ohms. 

Secondary  copper  loss  =  722r2  =  /i2ri  =  2,940  watts. 

Total  copper  loss  =  5,880  watts. 

Surface  of  each  coil  =  mean  turn  X  width  X  2  =  117.4  X  10.5  X  2 

=  2,470  sq.  in. 

Area  covered  by  spacing  blocks  =  360  sq.  in.  (Fig.  397). 
Radiating  surface  per  coil  =  2,470  -  360  =  2,110  sq.  in. 
Radiating  surface  of  H.V.  winding  =  2,110  X  6  =  12,660  sq.  in. 

12  660 
Square  inches  per  watt  lost  =      QA      =  4-3. 


Square  inches  per  watt  lost  on  L.V.  winding  =  4.3. 
Total  watts  lost  =  3,877  +  5,880  =  9,757. 
,   Cooling  coil  surface  required  =  9,757  X  1.0  =  9,757  sq.  in. 


TRANSFORMERS  413 

Diameter  of  pipe  =  1.25  in. 

9  757 

Length  of  pipe  =  ^35  x'3.14  X  12  =  2°7  ft* 
Diameter  of  case  =  5  ft. 
Length  of  one  turn  of  cooling  coil  =  3.14  X  5  =  15.7  ft. 

207 
Number  of  turns  of  cooling  coil  =  y^  =  13  turns. 

Efficiency  at  full-load  unity  power  factor  =  i  Qoo'oob'-u  9  757 

X  100  per  cent.  =  99  per  cent. 

(540) 2  X  117  4       1 
Primary  reactance  =  xi  =  20.2  X  10~8  X  60  X j^s X  6 

(°f>+l)=3.7ohm, 
Primary  reactance  drcp  =  I&i  =  45.4  X  3.7  =  168  volts 


Secondary  reactance  =  X*  =  20.2  X  10~8  X  60  X  (18) 2  X      X 


(°f-5+')  =  0.035  ohms. 
Secondary  reactance  drop  =  72z2  =  454  X  0.035  =  15.9  volts 

2^jQ  X  100  =  0.72  per  cent. 
Total  reactance  drop  =  1.48  per  cent. 

ft    A      K 

Primary  resistance  drop  =  Itfi  =  45 .4  X  1.42  =  64.5  volts   =  22  QOQ 

X  100  =  0.294  per  cent. 
6.45 
Secondary  resistance  drop  =  72r2  =  454  X  0.0142  =  6.45  volts  = 


X  100  =  0.294  per  cent. 
Total  resistance  drop  =  0.588  per  cent. 

(1  48)2 
Regulation  at  unity  power  factor  =  0.588  +     2QQ     =  0.6  per  cent. 

Regulation  at  80  per  cent,  power  factor  =  0.588  X  0.8  +  1.48  X  0.6 

(1.48   X  0.8  -  0.588  X  0.6)2 
+  -  2QQ  ~  =  1-35  per  cent. 


CHAPTER  XI 


CONVERTERS 

339.  Types  of  Converters. — Converters  are  rotating  machines 
which  change  electrical  energy  from  one  form  to  another.  They 
are  of  several  types : 

1.  A  synchronous  converter  or  rotary  converter  converts  from 
an  alternating  to  a  direct  current  or  vice  versa. 

2.  A  frequency  converter  converts  the  power  of  an  alternating- 
current  system  from  one  frequency  to  another  with  or  without  a 
change  in  the  number  of  phases  or  in  the  voltage. 

3.  A  rotary  phase  converter  converts  from  an  alternating-cur- 
rent system  of  one  or  more  phases  to  an  alternating-current  sys- 
tem of  a  different  number  of  phases  but  of  the  same  frequency. 

The  most  usual  change  is  from 
single  phase  to  polyphase  (see 
Art.  414).  Changes  from  one 
polyphase  system  to  another  are 
usually  carried  out  by  means  of 
static  transformers  (Art.  318). 

Any  of  these  conversions  can 
be  made  with  motor-generator 
sets. 

340.  Synchronous  Converter. 
— The  synchronous  converter  or 
rotary  converter  is  a  combina- 
tion of  a  synchronous  motor  and  a  direct-current  generator.  It 
receives  alternating  current  and  converts  it  into  direct  current. 
The  fields  are  excited  by  a  shunt  winding  connected  between 
the  direct-current  brushes  or  from  a  separate  exciter.  In  some 
cases  a  series  winding  is  added  for  compounding. 

The  armature  winding  is  an  ordinary  direct-current  winding 
and  may  be  either  series  or  multiple.  It  is  connected  to  a  com- 
mutator and  taps  are  taken  out  from  it  at  equidistant  points  and 
connected  to  slip  rings.  The  alternating  current  is  delivered  to 
the  slip  rings  either  single-phase,  two-phase,  three-phase  or  six- 

414 


FIG.  399. — Ring-wound  rotary 
converter. 


CONVERTERS 


415 


phase  and  drives  the  armature  as  a  synchronous  motor.  The 
same  armature  conductors  generate  and  carry  the  direct  current. 

Fig.  399  shows  a  ring-wound  bipolar  armature  tapped  for 
single-,  two-,  or  three-phase  currents;  single-phase  1  to  2  or  3  to  4; 
two-phase  1  to  2  and  3  to  4;  three-phase  1  to  5,  5  to  6  and  6  to  1. 

Fig.  400  shows  a  six-circuit  multiple  winding  for  a  six-pole, 
three-phase  rotary  converter  and  Fig.  401  shows  a  two-circuit  or 
series  winding  for  an  eight-pole,  three-phase  converter. 


FIG.  400. — Multiple-drum  winding  for  a  three-phase  rotary  converter. 

In  a  series-wound  armature  the  total  number  of  coils  must  be 
divisible  by  the  number  of  phases  and  in  a  multiple- wound  arma- 
ture the  number  of  coils  per  pair  of  poles  must  be  divisible  by  the 
number  of  phases. 

Six-phase  converters  are  operated  from  three-phase  circuits. 
Practically  all  converters  above  200  or  300  kw.  output  are  now 
built  with  six  collector  rings  and  six  phases.  Three  methods  of 
connecting  the  supply  transformers  are  shown  in  Fig.  370. 

341.  Ratios  of  E.M.Fs.  and  Currents. — With  the  brushes 
fixed  on  the  no-load  neutral  line,  the  direct  and  alternating  e.m.fs. 
generated  in  the  converter  bear  a  definite  relation  to  each  other 
and  one  cannot  be  varied  without  varying  the  other.  At  unity 


416 


ELECTRICAL  ENGINEERING 


power  factor  the  alternating  and  direct  currents  in  the  armature 
also  bear  a  definite  relation  to  each  other  if  the  current  required 
to  supply  the  converter  losses  is  neglected. 

Since  the  alternating  and  direct  currents  flow  in.  the  same  arma- 
ture conductors  and  in  opposite  directions,  the  e.m.f .  consumed  in 
the  armature  is  small  and  the  power  loss  is  small.  In  the  follow- 
ing analysis  these  quantities  will  be  neglected  and  the  alternating 
current  will  be  assumed  to  be  in  phase  with  the  impressed  e.m.f. 


r 


FIG.  401. — Two-circuit,  retrogressive  winding  for  a  three-phase  rotary  converter. 

This  condition  can  be  obtained  in  practice  by  properly  adjusting 
the  exciting  current. 

Take  first  the  case  of  the  single-phase  converter,  Fig.  402. 

Let  E  =  direct  voltage  of  the  converter. 

/  =  direct-current  output. 
Ei  =  effective  value  of  the  alternating  supply  voltage, 

whether  single  phase  or  polyphase. 
1 1  =  alternating  current  in  the  supply  lines. 
/'  =  alternating  current  in  the  armature  winding. 

The  voltage  between  the  leads  li  and  Z2  or  between  the  slip 
rings  Ri  and  R2  is  alternating  and  reaches  its  maximum  value 


CONVERTERS 


417 


when  li  and  12  are  under  the  brushes  and  it  is  then  equal  to  the 
direct  voltage  of  the  machine.  Therefore,  in  a  single-phase  con- 
verter the  direct  voltage  is  equal  to  the  maximum  value  of  the 
alternating  voltage  and  thus 

E  =  \/2#i  (357) 

or 

E  (358) 


Neglecting  losses  and  phase  displacements  the  output  of  the  con- 
verter is  equal'to  the  volt-amperes  input  or 

77»T          7?  7" 
Jiil   -=  Hi  il\ 

and  the  alternating  current  in  the  line  is 

(359) 


The  alternating  current  in  the  winding  is 

r  =  -1, 

p 

where  p  is  the  number  of  circuits  in  multiple  through  the  winding. 
In  the  bipolar  machine  in  Fig.  402,  p  =  2  and,  therefore,  • 

$.*-•**.  ^ 


Fio.  402. — Single-phase  converter. 


FIG.  403. — Two-phase  or  quarter-phase 
converter. 


342.  Two -phase  or  Quarter-phase  Converter. — When  four 
collector  rings  Ri}  R2,  Rs  and  R^,  Fig.  403,  are  connected  to  four 
equidistant  points  li,  12,  h  and  Z4,  the  machine  is  a  two-phase  or 
quarter-phase  converter.  The  two  voltages-  Ri  to  R2  and  Rs  to 
R±  are  equal  and  are  in  quadrature,  forming  a  two-phase  system; 

27 


418 


ELECTRICAL  ENGINEERING 


the  four  voltages  RI  to  R3)  R3  to  R2,  Rz  to  R±  and  R4  to  Ri  are 
all  equal  and  form  a  four-phase  or  quarter-phase  system. 

The  voltage  between  lines  or  the  voltage  per  phase  of  the  two- 
phase  supply  is  E 

"' 


The  voltage  between  adjacent  rings  or  the  quarter-phase  vol- 
tage is  Ei        E 

' 


Assuming  the  volt-amperes  input  two-phase  to  be  equal  to  the 
direct-current  output,  that  is, 

2E1Il  =  EI, 
the  alternating  current  per  line  is 

77T  T  77T  T  T 

T  •&•«  •&*  1  foao\ 

7i  =  ^r        —jjr       —7£  (363) 


The  alternating  current  in  the  winding  is  the  resultant  of  two 

currents  —  =        ,-  in  quadrature  and  its  value  is  therefore 

2       2\/2 


F  = 


(364) 


343.  Three-phase  Converter.  —  With  three  collector  rings  Ri, 
R2  and  R3,  Fig.  404,  connected  to  three  equidistant  points  h,  12 
and  £3  the  machine  is  a  three-phase  converter. 


Fro.  404. — Three-phase  converter.     FIG.  405. — E.m.fs.  and  currents 

in  a  three-phase  converter. 

.The  e.m.f.  between  each  of  the  rings  and  the  neutral  point  or 
the  "star"  e.m.f.  is  equal  to  half  of  the  single-phase  voltage.  It 
is  shown  as  Es  in  Fig.  405. 


CONVERTERS 
The  e.m.f.  between  rings  or  "  delta"  e.m.f.  is 

Ei  =  V3ES  =  ^!  =  0.612E. 
2\/2 

The  power  input  is 

i  =  3^Ji  =  3EJ' 


419 


(366) 


and  is  equal  to  the  output  EI. 
Thus  the  line  current  is 


and  the  alternating  current  in  the  winding  is 


I'  =      L  =  j  =  0.5457. 

A/3       3V3 


(368) 


FIG.  406.  —  n-phase  converter. 


FIG.  407. 


344.  n-phase  Converter.  —  For  an  n-phase  converter,  Fig.  406 
the  winding  must  be  tapped  at  n  equidistant  points.  The  e.m.f. 
between  each  of  the  rings  and  the  neutral  point  or  the  "star" 
e.m.f.  is  as  before 


The  e.m.f.  between  rings  or  the  e.m.f.  between  lines  is  the  vector 
difference  between  two  e.m.f  s.  Es  displaced  at  --  radians  (Fig. 
407).  Thus'.;. 


2Ea  sin  -  =  n 


The  power  input  is 

nEsIi  = 
and  is  equal  to  the  output  EI. 


,_ 
\/2 


(369) 


420 


ELECTRICAL  ENGINEERING 


Therefore  the  alternating  current  in  the  line  is 
£7  El       =  2\/2 

nEK  E  n       ' 


h  = 


n 


and  the  alternating  current  in  the  winding  is 
El  El  \/2I 


I'  = 


.      7T 

n  sin  - 
n 


(370) 


(371) 


The  values  obtained  above  for  the  e.m.fs.  and  currents  in  single- 
phase,  two-phase  and  three-phase  converters  can  also  be  ob- 
tained by  substituting  the  proper  values  of  n  in  equations  369, 
370  and  371 ;  single-phase  n  =  2,  two-phase  or  four-phase  n  =  4, 
and  three-phase  n  =  3.  The  results  are  tabulated  in  Fig.  408. 


Type 

Single- 
phase 

n  =  2 

Three- 
phase 
n  =  3 

Two-phase 
or  four- 
phase 
n  =  4 

Six-phase 
n  =  6 

n-phase 

E.m.f.   between   collector 

E 

VZE 

E 

E 

777       •        V 

JT,  gin  — 

rings  or  line  e.m.f.  E\.  .  . 

V2 

2V2 

2 

2-\/2 

n 

Current  per  line  I\  

V2/ 

2V2/ 

7 

V2/ 

V2 
2\/2/ 

3 

V2 

3 

n 

V2I 

Current  in  the  winding  /' 

I 

2\/2/ 

I 

V2/ 

,    if 

V* 

3\/3 

2 

3 

n  sin  - 
n 

FIG.  408. 

These  ratios  of  currents  only  hold  on  the  assumption  that  the 
power  factor  is  unity  and  that  the  efficiency  is  100  per  cent. 
The  power  factor  can  be  maintained  approximately  unity  by 
adjusting  the  excitation,  but  the  power  losses  cannot  be  elimi- 
nated and  the  values  of  alternating  current  in  the,  table  must  be 
increased  by  the  small  component  required  to  supply  the  losses  in 
the  machine.  When  the  power  factor  is  not  unity  the  reactive 
or  wattless  components  of  current  must  be  added  to  the  power 
components. 

The  ratios  of  e.m.fs.  are*  the  ratios  of  the  generated  e.m.fs. 
and  can  only  approximately  represent  the  ratios  of  terminal 


CONVERTERS 


421 


e.m.fs.  since  components  of  e.m.f.  are  consumed  in  the  resistance 
and  reactance  of  the  armature.  The  ratios  also  depend  on  the 
assumption  that  the  alternating  e.m.f.  wave  is  a  sine  wave.  If 
the  wave  is  peaked,  the  ratio  of  the  effective  value  to  the  maxi- 
mum value  is  less  than  —7=  and  the  values  of  the  alternating 
e.m.fs.  must  be  reduced.  If  the  wave  is  flat  topped  the  ratio  of 
effective  to  maximum  value  is  greater  than  —7=  and  the  values  of 
the  e.m.fs.  must  be  increased. 


FIG.  409. 

(2)  Alternating  Current 


(D- 


FIG.  410. — Current  in  coil  c,  Fig. 
409,  at  unity  power-factor. 


FIG.  411. — Current  in  coil  a,  Fig. 
409,  at  unity  power-factor. 


345.  Wave  Forms  of  Currents  in  the  Armature  Coils. — The 

current  in  any  armature  coil  is  the  difference  between  the  alter- 
nating-current input  and  the  direct-current  output. 

In  Fig.  409,  li  and  1%  are  the  two  leads  of  one  of  the  n-phases  of 
a  converter,  a  is  the  coil  next  to  one  lead  and  c  is  the  coil  in  the 
center  of  the  phase.  The  alternating  e.m.f.  and  the  power  com- 
ponent of  the  alternating  current  in  the  phase  li  to  12  are  maximum 
when  this  section  of  the  winding  is  midway  between  the  brushes 
and  they  are  both  zero  when  the  center  coil  c  is  under  the  brush. 

Fig.  410  shows  the  resultant  of  the  alternating  and  direct 


422 


ELECTRICAL  ENGINEERING 


current's  in  coil  c  during  one  revolution.  The  alternating  current 
is  opposed  to  the  direct  current  and  is  zero  when  the  direct  current 
reverses  as  the  coil  passes  under  the  brush.  The  current  in  the 
center  coil  is,  therefore,  less  than  the  current  in  any  other  coil 
in  the  phase  when  the  power  factor  is  unity. 

Fig.  411  shows  the  current  in  coil  a  next  to  one  of  the  leads. 
The  alternating  and  direct  currents  are  not  directly  opposing  and 
the  resultant  current  is  greater  than  in  the  center  coil  c. 

The  coils  next  to  the  leads,  therefore,  carry  larger  currents  than 
the  coils  in  the  center  of  the  phases  and  they  rise  to  a  higher 
temperature. 

The  worst  condition  of  local  heating  occurs  in  the  coil  next  the 
lead  of  a  single-phase  converter,  Fig.  412.  The  alternating  cur- 
rent has  its  maximum  value  when  the  direct  current  reverses. 


Alternating  Current 
© 


77 


® 


\\ 


Q)   Direct  Current 

Resultant  Current 


FIG.  412. — Current  in  the  coil  next  to 
the  lead  in  a  single-phase  converter. 


x — NS^— ©  Alternating  Currei 
\@  Kesultant  Current 

0 \    \    180 


\\ 


(TXDirect  CurrentX 


FIG.  413.— Current  in  coil  c,  Fig.  409, 
at  70  per  cent,  power  factor. 


When  the  power  factor  is  not  unity  the  minimum  resultant 
current  will  not  occur  in  the  center  coil  of  a  phase  but  in  a  coil 
displaced  from  it  by  the  angle  of  lag  or  lead  of  the  current.  Fig. 
413  shows  the  current  in  coil  c  when  the  power  factor  is  70  per 
cent,  and  a  component  of  lagging  current  equal  to  the  power 
current  flows  in  the  armature. 

346.  Heating  Due  to  Armature  Copper  Loss. — Take  the  case 
of  a  two-pole,  n-phase  armature,  Fig.  409,  and  use  the  same 
notation  as  before.  In  the  center  coil  c  of  the  phase,  the  direct 

current  is  ^  and  the  effective  value  of  the  alternating  current  is 


nsin  - 
n 


CONVERTERS 


423 


The  instantaneous  value  of  the  alternating  current  is 


i  =  A/2  F  sin  0  = 


21 


n  sin  - 

n 


sin  6, 


and  the  instantaneous  value  of  the  resultant  current  is 

27 


(372) 


(373) 


In  an  armature  coil  d  displaced  by  angle  0  from  the  center  of 
the  phase  the  alternating  current  is 

i  =  A/27'  sin  (0  -  0)  (374) 

and  the  instantaneous  value  of  the  resultant  current  is 

sin  (0  -  0)  -  75 


7T 

n  sin  - 
n 


4  sin  (0  -  0) 


-  1 


sin  - 

71 


(375) 


The  effective  value  of  the  resultant  current  is 


7     j  /v 

-^-JliW.  5   /-J,      nsi 


4  sin  (0  -  0) 


(376) 


n 


|ip 

TTjO 


16  sin2(0  -  0)  _  8  sin  (0  -  0)         1 

7T  /i 

n  sin  - 

71 


o      •     o  7T 

7i2  sin2  - 
n 


-  cos  2(0  -  0)(       8  sin  (0  -  0) 


+  1 


Jo[ 

o      •     o  If 

7i2  sin2  - 
n 

.    TT              \av 
n  sin  - 
n             J 

"       8 

sin  2 

(9  -  ft)  }    ,   8  cos  (.0  -  0)    ,    fl' 

0 

n2  sin2  - 

n 

2~           H 
7i  sm  — 
n 

["      8r 

16  cos  0 

~h  TT 

*• 

n2  sin2  - 

71 

.       7T 

7i  sm  - 
n 

16  cos 


(377) 


n  TT  sin 
n  n 


424  ELECTRICAL  ENGINEERING 

Since  ^  is  the  current  in  the  coil  when  the  machine  is  operating  as 

a  direct-current  generator,  the  ratio  of  the  power  lost  in  the  coil 
when  operating  as  a  converter  to  that  lost  when  operating  as  a 
direct-current  generator  with  the  same  output  is 

16  cos  ft  (37g) 


and  this  is  the  ratio  of  the  coil  heating  under  the  two  conditions. 
This  ratio  is  a  maximum  for  the  coil  next  to  the  alternating 


leads  h  or  Z2,  where  0  =  — ,  and  it  is 

IV 


«,      .      „  7T  .       7T 

n2  sin2  -       mr  sin  - 

n  n 


7T 

16  cos  - 

n  +  1.  (379) 


It  is  a  minimum  for  the  center  coil  of  the  phase,  where  (3  =  0, 
and  is 

h0  =  -  -^:  +  1.  (380) 

n2  sin2  —       mr  sin  - 
n  n 

The  ratio  of  the  total  power  lost  in  the  armature  of  the  con- 
verter to  that  lost  when  the  machine  is  operating  as  a  direct- 
current  generator  with  the  same  output  is  found  by  integrating 

the  ratio  h$  over  one-half  phase  from  (3  =  —  to  0  =  0  and  taking 
the  average.     It  is 

/»--  /£/       8  16  cos/3        A  ,. 


hftdp  =  - 

/o  I  n 

2  sin2 

7T 

7T 

ITT  sm  - 

r* 

n 

7T 

n\ 

8/3 

16 

sin  |8 

+  18 

f 

w\ 

n2  sin2  - 

mr 

7T 

sm  - 

\ 

n 

n 

Jo 

'  8 

16    , 

^ 

S  +  l,  (381) 

n2  sin2  - 
n 

and  this  is  the  relative  armature  heating  under  the  two  conditions. 
To  get  the  same  loss  in  the  armature  of  a  converter  and  the 


CONVERTERS 


425 


same  heating  as  in  the  direct-current  generator,  the  armature 

current  and  the  output  may  be  increased  in  the  ratio  ~~/~' 

V H 

The  values  of  H  and      /—  for  the  various  polyphase  converters 
are  tabulated  in  Fig.  414. 


Type 

Direct- 
current 
generator 

Single- 
phase 
n  =  2 

Three- 
phase 
n  =  3 

Two-phase 
or  four- 
phase 

n  =  4 

Six-phase 
n  =  6 

Relative   armature   heat- 
ing H 

1  00 

1  37 

0  55 

0  37 

0  26 

Rating  by  armature  heat- 
1 

me     /— 

1  .  00 

0.  85 

1  34 

1  64 

1  96 

ev# 

FIG.  414. 


For  a  single-phase  converter,  n  =  2,  the  value  of     ,—  is  0.85, 

and,  therefore,  the  output  of  a  machine  as  a  single-phase  converter 
is  only  85  per  cent,  of  its  output  as  a  direct-current  generator  for 
the  same  temperature  rise. 

For  a  three-phase  converter  — 7=  =  1.34  and  therefore  the  out- 
put is  34  per  cent,  greater  than  as  a  direct-current  generator. 
For  a  six-phase  converter  the  output  is  increased  96  per  cent. 

These  values  only  hold  if  the  alternating  current  is  in  phase 
with  the  impressed  e.m.f.  When  leading  or  lagging  currents  flow 
in  the  armature  the  heating  is  very  largely  increased  and  the  rating 
must  be  decreased. 

The  reduction  in  rating  due  to  reactive  lagging  or  leading  cur- 
rents may  be  found  as  follows : 

If  the  alternating  current  lags  behind  the  voltage  by  an  angle 
0,  then,  since  the  ratio  of  the  power  components  of  the  alternating 
and  direct  currents  must  be  the  same  as  before,  the  alternating 
current  input  is  increased  by  the  amount  of  the  reactive  lagging 
current  and  its  effective  value  is 


/'  = 


n  sin  -  cos  <j> 
n 


(382) 


426  ELECTRICAL  ENGINEERING 

and  the  instantaneous  value  of  the  resultant  current  in  the  coil 
d  is 

27  / 

sin  (0  -  |8  -  <f>)  -  -;  (383) 


\v  *r/  rt 

n  sin  -  cos  <f> 

n 

its  effective  value  is 

ZZJL —  _ iBeoBjfjE^.  (384) 

n2  sin2  -  cos2  </>  ?ITT  sin  - 

n  n 

the  ratio  of  the  power  lost  in  the  coil  of  a  converter  to  that  of  a 
direct-current  generator  is 

8  16cos(0+0)+  1(38g) 


n2  sin2  -  cos2  <£        WTT  sin  -  cos 
2  J  n  n 

and  the  relative  armature  heating  is 


n2  sin2-  cos2 


(386) 


To  find  the  power  factor  at  which  the  rating  of  the  converter 
is  reduced  to  that  of  the  direct-current  generator,  equate  H  to 
unity  and  solve  for  cos  <£. 

Thus, 

~  +  1  =  1  and  cos  </>  =  -  —  .   (387) 

n2  sin2  -  cos2  <f>  \/2n  sin  - 

n  n 

When  n  =  3,  cos  <f>  =  0.85  and  when  n  =  6,  cos  <£  =  0.745; 
therefore  the  rating  of  a  machine  as  a  three-phase  converter  will 
be  the  same  as  when  operated  as  a  direct-current  generator  when 
the  power  factor  is  85  per  -cent,  and  the  rating  of  a  machine  as  a 
six-phase  converter  will  be  the  same  as  when  operated  as  a  direct- 
current  generator  when  the  power  factor  is  about  75  per  cent. 

Synchronous  converters  should  be  operated  at  approximately 
unity  power  factor  at  full  load  and  overload. 

347.  Armature  Reaction.  —  The  armature  reaction  of  a  rotary 
converter  is  the  resultant  of  the  armature  reactions  of  the  machine 
as  a  direct-current  generator  and  as  a  synchronous  motor.  The 
direct-current  brushes  are  usually  placed  on  the  no-load  neutral 
points  and  therefore  the  direct-current  exerts  a  m.m.f.  in 


CONVERTERS 


427 


quadrature  behind  the  field  m.m.f.  (Fig.  415).  The  power  com- 
ponent of  the  armature  current  in  a  synchronous  motor  exerts 
a  m.m.f.  in  quadrature  ahead  of  the  field  m.m.f.  and  it  is  there- 
fore opposed  to  the  m.m.f.  of  the  direct  current. 

If  Z  is  the  number  of  conductors  on  the  armature  of  a  bipolar 

generator  and  ^-  is  the  direct  current  in  each  conductor,  the  arma- 

z 

Z  I 

ture  m.m.f.  is  the  resultant  of  ^  m.m.fs.  of  magnitude  ~  uni- 
formly distributed  over  the  circumference  of  the  armature  and 


Chord 


FIG.  415. — M.m.fs.  in  a  direct-current  generator. 


FIG.  417. 


it  is  therefore  less  than  that  of  a  concentrated  winding  in  the 

2 
ratio  -  (Fig.  416).     The  m.m.f.  of  the  direct  current  in  the  arma- 

7T 

ture  of  a  converter  is  thus 


Z    I    2     -  ZI 


(388) 


in  quadrature  behind  the  field  m.m.f. 

If  the  machine  is  connected  as  an  n-phase  converter,  the  num- 

9 

ber  of  turns  per  phase  is  ~-  and  the  effective  value  of  the  alternat- 
ing current  in  each  is 


/'  = 


V2I 

n  sin  - 
n 


428 


ELECTRICAL  ENGINEERING 


and  the  m.m.f.  per  phase  in  effective  ampere-turns  is 

ZT_  *  ZI 

2n  ~~ 


m' 


(389) 


These  ampere-turns  are  distributed  over  -th  of  the  circumfer- 
ence of  the  armature  and  their  resultant  is,  Fig.  417, 

r»         •       ^ 

2r  sin  - 


m  = 


ZIf    chord 

2n  '    arc 


ZT 

2n 


417), 


ZI 


r*     •    if 
2r  sin  - 


.      7T 

sin- 
n 


27TT 

n 


ZI 


The  maximum  value  of  the  m.m.f.  per  phase  is 

/«          Z/ 

mm  =  V  2m  =  —  ampere-turns. 

7m 


(390) 


(391) 


To  find  the  resultant  m.m.f.  of  the  alternating  current  in  the 
armature  it  is  necessary  to  combine  n  m.m.fs.   of  maximum 

ZI  2ir 

value  mm  =  -  -  displaced  in  direction  by  angle  —  and  displaced 

TT/t'  TL 

in  phase  by  -th  of  a  period  or  by  angle  —  • 

In  Fig.  418  phase  1  is  shown  in  the 
position  of  maximum  m.m.f.  if  the 
power  factor  is  unity.  The  direction  of 
the  m.m.f.  is  OB  and  it  is  in  quadrature 
ahead  of  the  field  m.m.f.  If  time  and 
angular  displacement  are  measured  from 
OB>  then  at  time  *  and  angle  e  the  m.m.f. 
of  phase  1  is  mm  cos  6  in  direction  OB\ 
and  its  component  in  direction  OB  is  mm  cos2  6. 

At  time  t  the  m.m.f.  of  phase  2  is  mm  cos  10  -\  --  j  making  an 
angle  (0  -\  --  )  with  OB  and  its  component  in  direction  OB  is 

\  77»  / 


Pioi8.Synchronou8 
motor. 


CONVERTERS  429 

The  resultant  m.m.f.  of  the  n  phases  in  the  direction  OB  at 
any  time  t  is 


Ma  =  mm    COB»  0+coB»  e  +  ~   +  +  +  cos 


=  ww  X  n  X  average  (cos)2  =  ram  = 

since  the  average  cos2  is  =  J^. 

The  resultant  m.m.f.  of  the  n  phases  in  the  direction  at  right 
angles  to  OB  or  in  line  with  the  field  m.m.f.  is  zero. 

Therefore  the  resultant  m.m.f.  of  the  alternating  current  in 
the  converter  armature  has  a  constant  value 

,,    .         n      ZI    n       ZI  /or^x 

^  =  m"2=™-2  =  27  (392) 

and  is  in  quadrature  ahead  of  the  field  m.m.f.  It  is  thus  equal  to 
the  m.m.f.  of  the  direct  current  and  is  opposed  to  it  and  the  re- 
sultant armature  reaction  of  the  direct  current  and  of  the  cor- 
responding power  component  of  the  alternating  current  is  zero. 

The  armature  reaction  due  to  the  power  current  required  to 
supply  the  losses  remains  but  it  is  very  small  and  produces  only 
a  slight  distortion  of  the  flux  in  the  air  gap. 

The  effective  armature  reaction  of  a  six-phase  converter  is 
from  7  to  20  per  cent,  of  that  of  the  corresponding  direct-current 
generator. 

When  the  power  factor  is  not  unity  the  wattless  currents  in 
the  armature  exert  m.m.f  s.,  as  in  the  synchronous  motor,  which 
act  in  line  with  the  field  m.m.f.  and  are  either  magnetizing  or 
demagnetizing  but  are  not  distorting. 

Thus  in  the  rotary  converter  there  is  very  little  field  distortion 
or  very  little  shifting  of  the  neutral  points  under  load.  As  a 
result  the  limit  of  overload  set  by  commutation  is  much  higher 
than  in  the  direct-current  generator.  This  is  very  important  in 
the  case  of  converters  supplying  railway  loads  where  the  load 
factor  is  usually  below  50  per  cent.  The  overload  capacities  for 
short  periods  must  be  high.  In  some  cases  when  using  interpoles 
momentary  overloads  of  200  per  cent,  are  permitted. 

348.  Control  of  the  Direct-current  Voltage.  —  Since  the  ratio  of 
alternating-  to  direct-current  voltage  is  fixed,  for  a  given  flux 
distribution,  the  direct  voltage  cannot  be  controlled  as  simply 
as  in  the  direct-current  generator.  There  are  three  practicable 
methods  of  control: 


430  ELECTRICAL  ENGINEERING 

(a)  Variation  of  the  alternating  voltage. 

(6)  Variation  of  the  direct-current  voltage  by  means  of  a 
direct-current  booster. 

(c)  Variation  of  the  flux  distribution  as  in  the  split-pole 
converter. 

349.  Methods  of  Varying  the  Alternating  Voltage.— The  im- 
pressed alternating  voltage  may  be  varied:  (1)  by  variable  ratio 
supply  transformers;  (2)  by  induction  regulators  in  the  supply 
lines;  (3)  by  introducing  reactances  in  the  supply  lines  and  draw- 
ing lagging  or  leading  currents  through  them  by  varying  the 
field  excitation  of  the  converter ;  and  (4)  by  a  synchronous  booster. 

The  first  method  has  the  -disadvantage  of  a  step-by-step 
regulation  and  the  contacts  are  liable  to  be  burned  as  the  circuit 
is  opened  when  changing  from  one  tap  to  another;  the  second 
method  requires  expensive  apparatus  but  gives  very  good  regula- 
tion and  may  be  made  automatic;  the  third  is  inexpensive  and 
is  made  automatic  by  placing  a  series  winding  on  the  converter 
poles.  The  fourth  method  requires  the  use  of  an  additional 
synchronous  machine.  It  is  very  satisfactory  and  is  being  em- 
ployed very  extensively  especially  in  large  installations. 

350.  Compounding  by  Reactance. — When  the  field  current  of 
a  synchronous  converter  is  varied,  reactive  lagging  or  leading 
currents  flow  in  the  armature  and  magnetize  or  demagnetize  the 
field  and  bring  it  back  to  its  former  strength  and  the  direct- 
current  voltage  remains  as  before.     To  vary  the  direct-current 
voltage  the  impressed  alternating  voltage  must  be  varied.     This 
may  be  accomplished  by  introducing  reactance  coils  in  the  supply 
lines  and  placing  a  series-field  winding  on  the  converter  poles. 
The  e.m.f.  of  inductance  lags  90  degrees  behind  the  current; 
thus,  when  the  converter  is  under-excited  a  component  of  current 
lagging  90  degrees  behind  the,  impressed  e.m.f.  flows  through 
the  reactance  coils,  and  the  e.m.f.  of  inductance  due  to  it  lags 
180  degrees  behind  the  impressed  e.m.f.  and  therefore  opposes 
and  lowers  it.     When  the  converter  is  over-excited  and  a  com- 
ponent of  current  90  degrees  ahead  of  the  e.m.f.  flows  through 
the  reactance,  the  e.m.f.  of  inductance  due  to  it  is  in  phase  with 
the  impressed  e.m.f.  and  raises  it. 

Therefore,  when  reactance  coils  are  connected  in  the  supply 
lines  an  increase  of  the  field  excitation  raises  the  impressed  e.m.f. 
and  so  raises  the  direct-current  voltage,  and  a  decrease  of  the  ex- 
citation lowers  the  impressed  e.m.f.  and  also  the  direct-current 


CONVERTERS 


431 


voltage.  The  result  is  the  same  as  in  the  direct-current  generator 
but  it  is  produced  in  a  somewhat  different  way. 

The  series  winding  causes  the  direct-current  voltage  to  rise 
automatically  with  load.  The  excitation  is  so  adjusted  that  at 
light  load  the  converter  is  under-excited  and  the  power  factor  is 
low  and  lagging  while  at  full  load  it  operates  at  a  power  factor  of 
about  98  per  cent,  leading.  This  results  in  only  a  slight  increase 
in  the  heating. 

In  some  cases  the  required  reactance  may  be  provided  in  the 
step-down  transformers  but  in  25-cycle  systems  it  is  difficult  to 
provide  sufficient  reactance  in  this  way. 


Alternating 


Booster 

Rheostat 

FIG.  419. — Synchronous  booster  converter. 

351.  Synchronous  Booster  Converter. — The  synchronous  boos- 
ter converter  is  a  combination  of  an  ordinary  converter  and  an 
alternating-current  generator  on  the  same  shaft  with  the  same 
number  of  poles  and  having  its  armature  winding  connected  in 
series  with  the  converter  armature.     The  excitation  of  the  booster 
is  so  arranged  that  it  can  be  reversed,  and  therefore  the  booster 
voltage  reversed,  giving  a  uniform  voltage  variation  of  double 
the  booster  voltage.     The  booster  may  be  shunt-excited  and  have 
its  voltage  controlled  •  by  an  automatic  regulator  or  it  may  be 
series- wound.     Fig.  419  shows  the  diagram  of  connections  for  a 
shunt-excited  booster. 

352.  Direct-current  Booster  Converter. — The  direct-current 
voltage  can  be  varied  by  inserting  a  booster  in  the  direct-current 


432 


ELECTRICAL  ENGINEERING 


(a)    Menu    do   Voltage 


(b)  Maxinaun^  j2,c  Voltage 


_ 

(c)  Mioimum   do  Voltage 


circuit.  It  may  be  direct-connected  to  the  converter  or  driven 
by  a  separate  motor.  In  either  case  additional  floor  space  is 
required  and  the  commutator  of  the  booster  must  have  the  same 
current  capacity  as  that  of  the  converter. 

353.  Split-pole   Converter. — In  the  split-pole   converter  the 
variation  of  direct-current  voltage  is  secured,  not  by  a  variation 
of  the  impressed  alternating  voltage  but  by  changing  the  shape 

I .        of  the  magnetic  field  so  that  the  total  flux 

/  \       and  the  direct-current  voltage  are  changed 

/  \     while  the  alternating  voltage  remains   un- 

changed. This  is  possible  because  in  three- 
phase  windings  third  harmonics  and  their 
multiples  do  not  appear  in  the  terminal 
voltage. 

If  the  pole  is  made  in  three  parts  the  exci- 
tation of  the  various  parts  may  be  changed 
so  as  to  introduce  a  third  harmonic  in  the 
flux  wave.  With  the  three  sections  equally 
excited  the  flux  wave  is  flat  as  shown  at  (a) 
in  Fig.  420  and  the  direct-current  voltage 
has  its  mean  value.  If  the  excitation  of  1 
and  3  is  increased  while  that  of  2  is  de- 
creased a  third  harmonic  is  produced  in  the 
flux  wave  and  in  the  generated  voltage  but 
will  not  appear  in  the  terminal  voltage  of  the 
three-phase  converter  with  its  taps  at  120  degrees.  The  total 
flux  and  therefore  the  direct-current  voltage  is  increased  in  pro- 
portion to  the  area  under  one-half  wave  of  the  third  harmonic 
flux  (Fig.  420  (6)).  If  the  excitation  of  1  and  3  is  decreased  and 
that  of  2  is  increased  the  flux  and  the  direct-current  voltage  are 
decreased  as  shown  at  (c).  A  variation  of  the  direct-current 
voltage  of  20  per  cent,  in  either  direction  may  be  obtained. 

The  space  required  for  the  three-part  pole  is  large  and  the  con- 
verter must  be  made  of  large  diameter  to  give  room  for  it. 
Similar  results  can  be  obtained  by  using  a  two-part  pole  but  the 
flux  wave  is  not  symmetrical  and  it  is  more  difficult  to  maintain 
a  suitable  commutating  field,  but  the  machine  has  a  smaller 
diameter  and  is  less  expensive. 

354.  Frequencies  and  Voltages. — Converters  are  built  for  25 
and  60  cycles  and  for  voltages  250  and  600  volts.     Some  25- 
cycle  converters  have  been  built  for  1,200  and  1,500  volts,  but 


Thrae  Part  Pole 

FIG.  420. — Split-pole 
converter. 


CONVERTERS  433 

where  such  high  direct-current  voltages  are  required  it  is  more 
usual  to  connect  two  converters  in  series  or  to  use  motor-genera- 
tor sets. 

The  design  of  converters  is  similar  to  that  of  direct-current 
generators  and  no  special  difficulties  are  met  in  25-cycle  conver- 
ters up  to  600, volts  but  in  60-cycle  converters  for  600  volts  and 
in  25-cycle  converters  for  1,200  volts,  since  the  number  of  poles 
is  fixed  by  the  frequency  and  the  speed,  it  is  difficult  to  leave 
space  enough  between  direct-current  brushes  for  a  sufficient 
number  of  commutator  bars.  The  bars  must  be  made  narrow 
and  the  number  of  volts  per  bar  is  high  and  the  danger  of  flash- 
over  is  great.  However,  by  increasing  the  commutator  periph- 
eral speed  to  5,500  ft.  per  minute  the  distance  between  neutral 
points  is  satisfactory  and  the  number  of  volts  per  bar  is  reduced 
to  a  safe  value.  Sixty-cycle  600-volt  converters  are  now  giving 
perfect  satisfaction  in  outputs  up  to  2,000  kw.  but  commutating 
poles  are  required  on  all  above  500  kw. 

355.  Outputs    and    Efficiencies. — Six-hundred-volt,    25-cycle 
converters  are  built  in  sizes  up  to  4,000  kw.  and  range  in  efficiency 
from  95  to  96  per  cent. ;  600-volt  60-cycle  converters  are  built  up 
to  2,000  kw.  and  with  efficiencies  from  91  to  94  per  cent.     Con- 
verters for  250  volts  are  built  up  to  1,000  kw.  with  efficiencies  of  93 
to  95.5  per  cent,  for  25  cycles  and  90.5  to  94  per  cent,  for  60  cycles. 

356.  Overload  Capacity. — Commutating-pole  converters  will 
stand  momentary  overloads  of  100  per  cent,  and  in  special  cases 
of  200  per  cent.     The  armature  reaction  is  much  smaller  than  in 
direct-current  generators  and  therefore  the  interpoles  do  not  re- 
quire so  much  excitation  and  can  be  designed  to  takei  ^are  of  very 
heavy  overloads  without  becoming  saturated. 

357.  Dampers. — Synchronous    converters   should    always   be 
provided  with  damper  windings  m  to  prevent  hunting.     A  com- 
plete squirrel-cage  winding  is  used  with  non-interpole  converters 
but  when  interpoles  are  required  separate  grids  are  placed  in  the 
main  pole  faces.     A  short-circuited  winding  surrounding  the  in- 
terpole  would  delay  the  growth  of  flux  in  it  and  so  interfere  with 
commutation. 

358.  Starting. — Synchronous  converters  may  be  started:  (1) 
from  the  alternating-current  end,  (2)  from  the  direct-current  end, 
and  (3)  by  an  auxiliary  induction  motor. 

359.  Alternating-current    Self-starting. — If    low    alternating 
voltage  is  applied  to  the  slip  rings  the  converter  will  start  as  an 

28 


434  ELECTRICAL  ENGINEERING 

induction  motor.  The  damper  windings  in  the  pole  faces  act 
as  the  secondary  and  the  rotating  armature  as  the  primary. 
When  it  is  nearly  up  to  synchronous  speed,  it  falls  into  step  as 
explained  in  Art.  277.  The  field  circuit  is  then  closed  and  full 
voltage  impressed.  For  small  converters  one  low-voltage  tap 
only  is  required  but  for  the  larger  ratings  two  starting  taps  are 
brought  out  from  the  supply  transformers. 

While  coming  up  to  speed  alternating  voltages  are  induced  in 
the  field  winding.  Their  magnitude  depends  on  the  ratio  of  field 
turns  to  armature  turns  and  on  the  slip  of  the  rotor  behind  the 
revolving  field,  and  they  disappear  when  synchronous  speed  is 
reached.  To  prevent  dangerous  induced  voltages  in  the  field 
winding  it  is  usually  broken  up  into  a  number  of  sections  during 
starting  by  means  of  a  field  break  up  switch,  or  it  may  be  short- 
circuited. 

The  converter  may  be  brought  up  to  speed  very  quickly  in  this 
way  since  it  does  not  require  to  be  synchronized,  but  there  is  no 
way  of  predetermining  the  polarity  of  the  direct-current  brushes. 
If  the  polarity  is  wrong  the  converter  may  be  forced  to  slip  a  pole 
by  reversing  the  field  switch  while  the  armature  is  still  connected 
to  the  low-voltage  taps.  The  switch  must  then  be  returned  to  its 
original  position. 

When  starting  from  the  alternating-current  end  a  large  current 
at  low  power-factor  is  drawn  from  the  lines  and  it  may  be  objec- 
tionable. 

360.  Direct-current  Self -starting. — Converters  may  be  started 
as  shunt  motors  from  the  direct-current  end  if  suitable  -power 
is  available  in  the  station  but  a  longer  time  is  required  to  put 
them  in  operation  than  with  alternating-current  starting  and 
they  must  be  synchronized.     The  starting  current  is  low. 

361.  Starting  by  an  Auxiliary  Motor. — If  an  induction  motor 
with  a  smaller  number  of  poles  than  the  converter,  and  conse- 
quently a  higher  synchronous  speed,  is  mounted  on  the  same 
shaft  it  may  be  used  as  a  starting  motor.     This  method  requires 
synchronizing  and  thus  takes  considerable  time  but  the  starting 
current  is  small  and  there  is  no  trouble  with  the  polarity.     The 
extra  starting  motor  increases  the  cost  of  the  equipment. 

362.  Brush  Lifting  Device. — When  starting  a  converter  as  an 
induction  motor  the  revolving  armature  flux  induces  voltages  in 
the  coils  short-circuited  by  the  direct-current  brushes  and  spark- 
ing results  but  it  is  not  usually  serious.     In  commutating-pole 


CONVERTERS  435 

converters  the  sparking  tends  to  be  very  much  worse  since  the 
reluctance  of  the  path  through  the  short-circuited  coil  is  reduced 
by  the  presence  of  the  interpole  iron  and  therefore  the  flux  and  the 
induced  voltage  are  increased.  To  prevent  serious  sparking  a 
brush-lifting  device  is  employed  which  raises  all  the  brushes 
except  two  narrow  ones  which  are  required  to  indicate  the 
polarity. 

363.  Bucking  or  Flashing. — Bucking  or  flashing  may  be  the  re- 
sult of  poor  commutating  conditions  causing  bad  sparking  under 
the  brushes,  or  of  high  voltage  between  adjacent  bars;  or  of  low 
surface  resistance  between  adjacent  brushes.     These  causes  are 
usually  all  present  to  a  greater  or  lesser  degree  when  flashing 
takes  place. 

Commutating  conditions  in  converters  are  normally  better 
than  in  direct-current  generators  since  the  resultant  armature 
reaction  is  small ;  but  very  heavy  overloads  may  distort  the  field 
and  cause  sparking ;  or  sparking  may  result  from  mechanical  de- 
fects as  wrong  brush  setting,  poor  contact,  high  mica,  etc. 

Comparatively  high  voltage  between  bars  is  the  usual  condition 
in  rotary  converters  since  the  space  between  neutral  points  is 
limited  especially  for  the  higher  frequencies  and  voltages. 

Any  increase  of  impressed  voltage  due  to  disturbances  on  the 
lines  will  increase  the  volts  per  bar  or  a  sudden  increase  of  cur- 
rent, due  to  a  short-circuit,  may  distort  the  field  and  increase  the 
voltage  between  certain  bars.  This  tends  to  cause  local  sparking 
between  these  bars.  A  spark  starting  under  the  brush  due  to 
any  cause  may  then  be  carried  round  to  the  next  brush  especially 
if  the  commutator  is  dirty.  Such  a  flash-over,  if  severe,  will 
short-circuit  the  converter  and  cause  it  to  buck,  fall  out  of 
synchronism  and  stop. 

364.  Parallel  Operation. — When  two  or  more  converters  are 
connected  to  the  same  direct-current  bus-bars  they  should  be  sup- 
plied from  separate  banks  of  transformers  to  prevent  large  cur- 
rents circulating 'between  them  due  to  differences  of  counter  e.m.f. 
or  to  variations  in  the  resistance  of  the  direct-current  brush  con- 
tacts.    The  separate  transformer  banks  permit  slight  corrections 
to  be  made  in  the  voltages  of  the  converters. 

365.  Inverted  Converter. — Where  a  small  alternating-current 
load  is  to  be  supplied  from  a  direct-current  system,  a  rotary  con- 
verter may  be  used  as  an  inverted  converter  to  transform  direct 
current  to  alternating  current. 


436  ELECTRICAL  ENGINEERING 

The  ratios  of  the  voltages  are  the  same  as  under  normal  oper- 
ating conditions  but  the  ratios  of  the  currents  vary  since  it  is  not 
possible  to  eliminate  or  control  the  wattless  components  of  the 
alternating  current.  These  components  depend  on  the  character 
of  the  load  and  are  not  affected  by  varying  the  exciting  current. 

When  changing  from  alternating  current  to  direct  current  the 
speed  of  the  converter  is  fixed  by  the  frequency  of  the  system  and 
remains  constant.  When  changing  from  direct  current  to  alter- 
nating current  the  speed  is  not  fixed  but  depends  on  the  excita- 
tion and  varies  as  the  field  strength  varies.  When  the  load  is 
inductive  the  wattless  lagging  current  demagnetizes  the  field  and 
so  raises  the  speed  of  the  converter  and  the  frequency  of  the  alter- 
nating current.  This  may  increase  the  lagging  current  and  so 
raise  the  speed  more  until  it  gets  beyond  safe  limits.  The  in- 
verted converter  has  thus  the  two  disadvantages:  (1)  that  it 
tends  to  run  at  dangerous  speeds,  and  (2)  that  it  supplies  a  cur- 
rent of  varying  frequency.  It  must,  therefore,  be  provided  with 
some  means  of  cutting  off  the  load  when  the  speed  rises  above  a 
certain  value  or  with  some  means  of  limiting  the  speed. 
'  If  the  converter  is  excited  by  a  direct-current  generator 
mounted  on  the  same  shaft  any  increase  in  speed  raises  the  exciter 
voltage  at  a  higher  rate  and,  therefore,  the  field  of  the  converter 
is  strengthened  and  the  speed  is  limited. 

Rotary  converters  to  be  used  as  inverted  converters  should  be 
shunt-wound  to  secure  as  constant  a  speed  and  frequency  as 
possible. 

If  two  converters  are  operating  in  parallel  and  the  alternating- 
current  circuit  breakers  of  one  open,  this  converter  will  run  in- 
verted as  a  direct-current  shunt  motor  and  if  it  is  under-excited 
the  speed  will  rise.  If  it  is  in  a  position  to  supply  alternating 
current  to  a  reactive  circuit  the  rise  in  speed  may  be  serious. 
This  condition  may  be  prevented  if  the  direct-  and  alternating- 
current  circuit-breakers  are  interlocked  or  if  the  direct-current 
breakers  are  provided  with  a  trip  operated  by  a  reverse-current 
relay. 

366.  Double-current  Generator. — If  mechanical  power  is  sup- 
plied to  drive  a  rotary  converter  it  can  be  used  as  a  double-current 
generator  to  supply  direct  current  from  the  commutator  and 
alternating  current  from  the  slip  rings. 

The  two  currents  in  this  case  flow  in  the  same  direction  in  the 
armature  conductors  and  the  losses  are  increased  and  the  arma- 


CONVERTERS 


437 


ture  reaction  is  the  sum  of  the  reactions  due  to  the  two  currents. 
The  voltage  regulation  is,  therefore,  poorer  than  in  the  converter. 

367.  Three-wire  Generator. — The  three-wire  direct-current 
generator  is  similar  in  construction  to  a  single-phase  or  quarter- 
phase  rotary  converter.  It  is  used  to  supply  a  three-wire  system 
with  from  220  to  280  volts  between  outer  wires  and  from  110  to 
140  volts  between  each  of  the  outer  wires  and  the  neutral  wire. 

In  order  to  obtain  a  point  at  a  potential  midway  between  the 
potentials  of  the  direct-current  brushes  special  transformers  called 
compensators  are  used.  They  have  a  single  winding  tapped  at 
the  center  and  are  connected  by  means  of  slip  rings  across  points 
on  the  armature  winding  180  electrical  degrees  apart  (Fig.  421). 
The  neutral  wire  of  the  system  is  connected  to  the  central  point 
of  the  compensator  and  its  potential  is  maintained  almost  mid- 
way between  the  potentials  of  the  outer  wires.  When  more  than 
one  compensator  is  used  the  center  points  of  all  the  compensators 
are  joined  together  before  being  connected  to  the  neutral  wire 
(Fig.  422). 


FIG.  421. — Three- wire  generator  with      FIG.  422. — Three- wire  generator  with 
a  single  compensator.  two  compensators. 

In  some  cases  the  compensators  are  connected  directly  to  the 
armature  windings  and  rotate  with  it  and  their  neutral  points  are 
connected  and  brought  out  to  a  single  slip  ring. 

The  voltage  between  the  termirials  of  the  compensators  is 
alternating  and  when  the  loads  on  the  two  sides  of  the  three-wire 
system  are  equal  and  no  current  flows  in  the  neutral  wire  the  only 
current  in  the  compensator  is  the  very  small  exciting  current. 

When  the  loads  are  unequal  the  unbalanced  current  IN  flows  in 
the  neutral  wire  as  shown  in  Fig.  421.  On  reaching  the  com- 
pensator it  divides  into  two  parts  which  flow  through  the  winding 
and  up  to  the  positive  brush  by  the  path  of  least  resistance. 
Since  the  current  in  the  neutral  wire  is  a  direct-current  the  react- 


438 


ELECTRICAL  ENGINEERING 


ance  of  the  compensator  does  not  oppose  it  and  the  only  voltage 
drop  is  that  due  to  resistance. 

The  actual  amount  of  current  carried  by  the  various  sections 
of  the  armature  winding  is  very  difficult  to  calculate  and  it  varies 
from  instant  to  instant  due  to  the  change  in  the  relative  positions 
of  the  direct-current  brushes  and  the  compensator  connections. 
The  average  current  carried  by  each  half  of  the  compensator 

.:...          '.'In 

winding  is  .-*• 

The  unbalancing  of  the  currents  in  the  sections  of  the  armature 
winding  produces  an  unbalancing  of  the  armature  reactions  and 
results  in  a  slight  unbalancing  of  the  voltages  between  the  neutral 
point  and  the  brushes.  With  two  or  more  compensators,  Fig. 
422,  this  unbalancing  is  reduced  due  to  the  more  even  distribu- 
tion of  the  current. 


Positive 


Neutral 

FIG.  423. — Parallel  operation  of  three-wire  generators. 


Machines  can  be  designed  to  give  a  regulation  of  2  per  cent, 
or  less  with  an  unbalanced  load  of  25  per  cent. 

With  this  system  the  voltages  on  the  two  sides  cannot  be  regu- 
lated independently  and  the  flexibility  of  the  three-wire  system 
supplied  by  t,wo  generators  in  series  is  lost,  but  there  is  a  corre- 
sponding gain  in  space  and  cost  of  machines. 

The  capacity  required  in  the  compensators  is  small.     With  25 


CONVERTERS  439 

per  cent,  unbalancing  of  the  loads  the  required  compensator 
capacity  is  less  than  10  per  cent,  of  the  generator  capacity. 

368.  Parallel  Operation  of  Three-wire  Generators. — Fig.  423 
shows  the  diagram  of  connections  for  parallel  operation  of  two 
compound-wound  three-wire  generators,  which  are  provided  with 
interpoles. 

The  series  windings  must  be  divided  into  two  groups,  one  on 
either  side  of  the  system,  and  two  equalizer  connections  are  re- 
quired. Half  of  each  interpole  winding  must  be  connected  on 
one  side  of  the  system  and  half  on  the  other.  In  the  case  of  the 
series-field  windings  it  is  sufficient  to  supply  the  south  poles  from 
one  side  and  the  north  poles  from  the  other. 

To  connect  No.  2  in  parallel  with  No.  1  bring  it  up  to  speed  and 
close  switches  Si,  Si,  thus  exciting  its  series  windings,  adjust  the 
voltage  by  means  of  the  shunt-field  rheostat  #,  close  $2,  Sz  and 
finally  close  the  neutral  switch  $3. 

Circuit-breakers  B,  B  with  overload  and  reverse-current  trip 
coils  should  be  connected  on  each  side  to  protect  the  machines. 

369.  Frequency  Converters. — Frequency  converters  are  used 
where  power  is  transmitted  at  25  cycles  and  is  required  by  the 
consumer  at  60  cycles  or  where  two  systems  of  different  fre- 
quencies are  to  be  linked  together  in  a  high-voltage  network. 
The  most  usual  form  of  frequency  converter  is  a  synchronous 
motor-generator  set,  a  25-cycle  motor  direct-connected  to  a  60- 
cycle  alternator.     The  numbers  of  poles  on  the  two  machines 
must  be  in  the  ratio  of  25  to  60.    When  a  10-pole  motor  is  used  the 
alternator  must  have  24  poles  and  the  speed  is  fixed  at  300  r.p.m. 

When  frequency  converters  are  to  be  operated  in  parallel  they 
must  be  synchronized  on  both  the  25-cycle  and  60-cycle  ends. 
If  the  motor  is  synchronized  first  there  is  only  one  chance  in  five 
that  the  alternator  is  in  synchronism,  while  if  the  alternator  is 
synchronized  first  there  is  only  one  chance  in  twelve  that  the 
motor  is  in  synchronism. 

If  the  motor  is  synchronized  and  it  is  found  that  the  generator 
is  out  of  synchronism  the  circuit  must  be  opened  and  the  motor 
allowed  to  slip  back  a  pair  of  poles  at  a  time  until  the  correct 
position  is  reached. 

When  linking  up  two  systems  by  a  frequency  converter,  the 
converter  must  be  connected  to  one  of  them  and  then  the  second 
system  brought  up  to  the  position  of  synchronism. 

The  induction  frequency  converter  is  discussed  in  Art.  415. 


440  ELECTRICAL  ENGINEERING 

370.  Mercury  Vapor  Rectifier. — A  rectifier  acts  as  an  electric 
valve  allowing  current  to  flow  through  it  in  only  one  direction 
and  it  may  therefore  be  used  to  rectify  an  alternating  current, 
that  is,  to  change  it  to  a  uni-directional  current. 

The  mercury  vapor  rectifier,  which  is  used  very  largely  to  con- 
vert alternating  current  to  direct  current  for  charging  storage 
batteries,  supplying  arc  lamps  and  many  other  purposes,  is  shown 
in  Fig.  424 (a).  It  consists  of  an  exhausted  bulb  B  which  has  two 
projections  on  its  sides  containing  the  positive  terminals  or  anodes 
A  and  A'  which  are  made  of  graphite,  and  two  projections  on  the 
bottom  containing  mercury ;  C  is  the  negative  terminal  or  cathode 
and  S  is  a  third  anode  used  only  for  starting.  The  large  upper 
space  in  the  bulb  is  the  cooling  chamber  in  which  the  mercury 
vapor,  which  has  been  heated  by  the  passage  of  electricity,  is 
condensed  and  from  which  it  falls  down  into  the  cathode  again. 

The  two  anodes  are  connected  to  the  terminals  of  the  trans- 
former TTf  and  the  load  circuit  is  connected  between  the  center 
point  of  the  transformer  and  the  cathode  C  in  series  with  the 
sustaining  coil  F. 

The  operation  of  the  rectifier  depends  on  the  fact  that  current 
can  pass  through  the  tube  in  one  direction  only,  from  either  of 
the  anodes  to  the  cathode,  and  it  can  only  pass  in  this  direction 
after  an  arc  has  been  formed  at  the  cathode  in  such  a  direction 
as  to  make  the  mercury  negative.  In  the  opposite  direction  the 
tube  is  a  very  good  insulator  and  a  difference  of  potential  of 
thousands  of  volts  would  be  required  to  produce  a  current.  The 
starting  arc  is  produced  by  impressing  a  voltage  between  the  two 
mercury  terminals  C  and  S  through  the  starting  resistance  rs 
and  tipping  the  tube  until  the  mercury  forms  a  bridge  and  closes 
the  circuit.  Current  then  passes  and  when  the  circuit  is  opened 
by  raising  the  tube  to  the  vertical  position  an  arc  is  formed  and 
the  cathode  is  said  to  be  excited.  If  at  this  instant  either  of  the 
anodes  is  at  a  higher  potential  than  the  cathode,  current  will 
flow  from  it  and  will  continue  to  flow  so  long  as  the  difference  of 
potential  is  greater  than  the  counter  e.m.f .  of  the  rectifier.  When 
the  terminal  T  of  the  supply  transformer  is  positive,  current 
flows  from  it  to  A  through  the  tube  to  C,  through  the  sustaining 
coil  F  and  load  circuit  to  the  terminal  0.  When  the  voltage 
reverses  and  T  becomes  negative,  T'  becomes  positive  and  cur- 
rent flows  from  it  to  A '  through  the  tube  to  C  and  through  the 
coil  F  to  the  terminal  0. 


CONVERTERS 


441 


If  there  were  no  drop  of  voltage  in  the  rectifier  the  current  wave 
would  be  the  same  shape  as  the  voltage  wave  and  in  the  load  cir- 
cuit it  would  vary  from  zero  to  a  positive  maximum.  There  is, 
however,  a  drop  of  voltage  of  from  14  volts  to  25  volts  in  the  com- 
mercial rectifier,  which  remains  approximately  constant  inde- 
pendent of  the  load,  and  if  the  sustaining  coil  were  left  out  of 
the  circuit  the  current  through  the  bulb  would  drop  to  zero  as 
soon  as  the  potential  of  T  had  fallen  below  the  counter  e.m.f.  of 
the  rectifier  and  load  circuit  and  would  remain  at  zero  until  the 
potential  of  Tr  rose  to  a  value  greater  than  this  counter  e.m.f. 
In  the  meantime  the  cathode  would  have  lost  its  excitation  and 


r 

n 

n 

V, 

L 

/ 

\ 

L 

\ 

L 

i 

\ 

y 

\ 

f 

\ 

r 

\ 

f 

\ 

^ 

^ 

"^ 

^ 

"^ 

•^-^ 

^x 

Impreesed 
0    Voltage 

Current  through 
Electrode   A 

Current  through 
-   Electrode   A' 

Rectified  Current 
in  the  Load  Circuit 

o 


Direction 
of  Current 


(a)  F  (6) 

FIG.  424. — Single-phase  mercury  vapor  rectifier. 


FIG.  425. — Three-phase 
mercury  vapor  rectifier. 


the  tube  would  have  to  be  tipped  again  before  any  current  could 
flow.  To  prevent  the  current  in  the  tube  from  falling  to  zero 
and  so  to  insure  continuous  operation  the  reactance  coil  F  is 
introduced.  Its  action  is  as  follows:  While  current  is  flowing 
from  T  energy  is  stored  in  the  magnetic  field  of  the  coil  and  when 
the  potential  of  T  becomes  too  low  to  force  the  current  against 
the  counter  e.m.f.  of  the  converter  the  coil  discharges  its  stored 
energy  and  maintains  the  current  until  the  potential  of  Tr  rises 
and  current  flows  from  it  to  the  load.  The  effect  of  the  sustain- 
ing coil  is  therefore  to  spread  out  the  two  halves  of  the  current 
wave  so  that  they  overlap  and  produce  in  the  load  circuit  a  direct 
current  with  only  a  slight  pulsation  (Fig.  424(6)). 

371.  Currents  and  Voltages. — The  voltage  is  controlled  by  a 
regulating  reactance  connected  in  the  alternating-current  supply 
circuit  and  in  the  ordinary  rectifiers  the  direct  voltage  ranges 
from  20  to  52  per  cent,  of  the  alternating  voltage  while  the  alter- 
nating current  ranges  from  40  to  66  per  cent,  of  tfte  direct  current. 

Rectifiers  are  designed  for  direct  currents  of  10,  20,  30  or  40 


442 


ELECTRICAL  ENGINEERING 


amp.  and  can  be  built  to  operate  on  any  required  voltage  and 
any  frequency. 

372.  Losses  and  Efficiency. — Since  the  counter  e.m.f.  of  the 
rectifier  is  approximately  constant  independent  of  the  load,  the 
power  losses  vary  directly  as  the  current  instead  of  as  the  square 
of  the  current.     Neglecting  the  losses  in  the  sustaining  coil  and 
regulator  the  efficiency  of  the  rectifier  is  constant  at  all  loads  and 
is  higher  the  higher  the  voltage.     Values  up  to  80  per  cent,  are 
reached  with  rectifiers  supplied  from  a  220-volt  alternating-cur- 
rent circuit  and  delivering  110  volts  direct  current. 

The  power  factor  of  the  rectifier  is  high  and  under  ordinary 
conditions  may  be  assumed  to  be  about  90  per  cent. 

373.  Three-phase  Rectifier. — The  three-phase  rectifier,  Fig. 
425,  does  not  require  any  sustaining  coil  since  the  voltage  waves 
of  the  three  phases  overlap  and  there  is  therefore  no  tendency  for 
the  load  current  to  fall  to  zero  and  allow  the  cathode  to  lose  its 
excitation.     It  must,  however,  be  started  in  the  same  way  as  the 
single-phase  rectifier. 


FIG.  426. — Hot-cathode  half-wave  rectifier. 


FIG.  427. — Hot-cathode  full- 
wave  rectifier. 


374.  Hot-cathode  Argon-filled  Rectifier. — Another  type  of 
rectifier  suitable  for  low-voltage  large-current  circuits  has  been 
developed  recently,  which  depends  on  the  principle  that  current 
can  flow  from  a  cold  to  a  hot  electrode  but  not  in  the  reverse 


CONVERTERS  443 

direction  (Fig.  426).  The  rectifier  consists  of  a  glass  bulb  filled 
with  argon  gas  with  a  graphite  anode  mounted  on  a  heavy 
tungsten  lead  and  two  tungsten  cathodes.  One  cathode  A  is 
used  for  starting  and  the  other  B  for  operating.  Cathode  A  is 
similar  to  the  filament  of  a  tungsten  lamp  and  is  excited  from  a 
40-watt  transformer  which  raises  it  to  a  very  high  temperature 
and  heats  the  operating  cathode  B.  If  the  bulb  is  connected  in 
series  with  a  battery  or  other  load  across  an  alternating-current 
supply  only  one-half  of  each  alternating  wave  can  pass.  If 
required,  two  half-wave  rectifiers  may  be  combined  in  one,  Fig. 
427,  and  both  halves  of  the  alternating  wave  passed  through  to 
the  battery.  Fig.  426(6)  and  Fig.  427(6)  show  the  methods  of 
connecting  the  half-wave  and  full-wave  rectifier  in  the  circuit. 
The  current  may  be  regulated  by  a  resistance  in  series  as  shown 
or  by  means  of  an  auto-transformer  in  the  supply  circuit. 

In  rectifiers  for  very  small  currents  it  is  necessary  to  keep  the 
exciting  cathode  in  operation  at  all  times  but  with  larger  currents 
it  is  only  required  for  starting.  When  the  exciting  cathode  is 
used  continuously  the  loss  in  it  and  in  the  supply  transformer  is 
offset  by  a  low  energy  loss  in  the  arc  between  the  electrodes.  The 
arc  drop  in  this  case  is  from  4  to  8  volts  while  without  the  exciting 
electrode  it  is  10  to  14  volts. 

The  rectifier  will  start  on  voltages  as  low  as  20  volts,  and  it 
is  used  principally  for  charging  small  storage  batteries. 


CHAPTER  XII 
INDUCTION  MOTOR 

375.  Induction  Motor. — Fig.  428  shows  an  induction  motor 
of  the  squirrel-cage  type.  It  consists  of  two  main  parts,  the 
primary  or  stator  and  the  secondary  or  rotor.  The  stator  is 
exactly  similar  to  the  armature  of  a  synchronous  motor.  The 
rotor  which  takes  the  place  of  the  rotating  field  member  of  the 
synchronous  motor  is  not  excited  by  direct  current  but  has  cur- 
rents induced  in  it  by  transformer  action  from  the  stator;  thus 


FIG.  428. — Induction  motor. 

the  transfer  of  power  from  the  stator  to  the  rotor  is  similar  to 
the  transfer  of  power  from  the  primary  to  the  secondary  in  a 
transformer.  The  rotor  is,  however,  free  to  move  and  there  is  an 
air  gap  in  the  magnetic  circuit  and  therefore  the  magnetizing 
current  is  large,  the  leakage  reactances  are  large  and  the  power 
factor  is  low. 

376.  The  Stator. — The  primary  or  stator  consists  of  a  winding 
carried  in  slots  on  the  inner  face  of  a  laminated  iron  core.     The 

444 


INDUCTION  MOTOR 


445 


winding  is  similar  to  an  alternator  or  synchronous  motor  winding 
and  the  coils  are  connected  in  groups  according  to  the  number  of 
phases  and  poles,  one  group  per  phase  per  pair  of  poles. 

The  stator  is  supplied  with  polyphase  alternating  currents  and 
a  revolving  m.m.f.  is  produced  similar  to  the  m.m.f.  of  armature 
reaction  in  an  alternator,  which  produces  a  magnetic  field  re- 
volving at  a  constant  speed  called  the  synchronous  speed  of  the 
motor. 

In  Fig.  429,  (a)  represents  the  stator  winding  of  a  two-pole, 
two-phase  induction  motor,  (6)  the  currents  supplied  to  the  two 


Phase  1,  z,=ImCo80    , 
Current       /          Phase  2  »2=Imsin  0 


(2)  (3)  (4) 

FIG.  429. — Revolving  m.m.f.  and  flux  in  a  two-pole,  two-phase  induction  motor. 


phases  and  (c)  the  fluxes  produced  by  the  resultant  stator  m.m.f. 
at  the  points  (1),  (2),  (3)  and  (4)  of  the  cycle. 

The  windings  start  at  Si  and  sz  and  finish  at  /i  and  /2  respec- 
tively. A  positive  current  is  one  which  enters  at  Si  or  s2  and  a 
negative  current  is  one  which  enters  at  f\  or  /2. 

Referring  to  Fig.  429  (c)  it  is  seen  that  the  north  pole  makes  one 
complete  revolution  in  the  anti-clockwise  direction  while  the 
current  in  phase  1  passes  through  one  cycle. 

Fig.  430  (a)  represents  the  stator  of  a  two-pole  -three-phase 
induction  motor,  430(6)  the  currents  supplied  and  430(c)  the 
fluxes  corresponding  to  the  points  (1),  (2),  (3)  and  (4)  on  the 
cycle.  The  north  pole  as  before  makes  one  revolution  during 
one  cycle. 


446 


ELECTRICAL  ENGINEERING 


Fig.  431  (a)  represents  the  stator  of  a  four-pole,  two-phase 
induction  motor,  (&)  the  currents  supplied  and  (c)  the  fluxes 
produced. 

Phases,  Z3=Im  Cos  (0-340) 
Cos  (0-120) 


FIG.  430. — Revolving  m.m.f.  and  flux  in  a  two-pole,  three-phase  induction  motor. 


(2)  (3)  (4) 

CO 

FIG.  431. — Revolving  m.m.f.  and  flux  in  a  four-pole,  two-phase  induction  motor. 

Fig.  432 (a),  (6)  and  (c)  represent  a  similar  set  of  conditions 
for  a  four-pole,  three-phase  stator. 

In  Fig.  431  (c)  and  Fig.  432(c)  it  is  seen  that,  while  the  current 


INDUCTION  MOTOR 


447 


goes  through  one  cycle,  the  revolving  field  turns  through  the  angle 
occupied  by  one  pair  of  poles. 

If  the  stator  winding  has  p-poles,  the  revolving  field  turns 

SfiO  2 

through  —T~  degrees,  that  is,  through  -  of  one  revolution  during 

one  cycle  of  the  current. 

If  the  frequency  of  the  supply  is  /  cycles  per  second  the  revolv- 

2f                120f 
ing  field  makes  —  r.p.s.  or r.p.m.     The  synchronous  speed  of 

an  induction  motor  is,  therefore, 


AT  120/ 

N  =  -      -  r.p.m. 
P 


(393) 


FIG.  432. — Revolving   m.m.f.    and   flux  in   a  four-pole,    three-phase  induction 

motor. 


377.  Revolving  Magnetomotive  Force  and  Flux  of  the  Stator. 
— In  Fig.  433,  OX  is  the  direction  of  the  m.m.f.  of  phase  1  of  the 
two-phase  motor  in  Fig.  429  and  its  value  at  any  instant  is 

mi  =  njm  sin  (0  +  90)  =  njm  cos  6, 

^  where  HI  is  the  number  of  turns  per  phase  and  ii  =  Im  sin  (0  + 
90)  is  the  current  in  phase  1. 

At  the  same  instant  the  m.m.f.  of  phase  2  is 

m2  =  nilm  sin  0,  in  direction  OF, 
where 

*2  =  Im  sin  0  is  the  current  in  phase  2. 


448 


ELECTRICAL  ENGINEERING 


The  resultant  m.m.f.  of  the  two  phases  is 


m  = 


+  m22  =  njm  Vcos2  0  +  sin2  0  = 


and  makes  an  angle  6  with  the  OX  axis. 

The  resultant  m.m.f.  is,  therefore,  constant  in  value,  being 
equal  to  the  maximum  m.m.f.  of  one  phase,  and  it  revolves  at 
synchronous  speed  in  the  anti-clockwise  direction. 


FIG.  433. 


FIG.  434. 


This  constant  m.mj.  acting  on  a  path  of  constant  reluctance 
produces  a  field  of  constant  strength  revolving  with  the  m.m.f. 
and,  therefore,  revolving  at  synchronous  speed  relative  to  the 
winding  of  the  stator.  The  flux  linking  with  each  phase  of  the 
stator  is  an  alternating  flux  which  reaches  its  maximum  when 
the  current  in  the  phase  is  maximum  and  is  therefore  in  phase 
with  it. 

Figs.  430 (a),  (b)  and  (c)  represent  respectively  the  winding  of  a 
three-phase,  two-pole  stator,  the  currents  supplied  and  the 
m.m.fs.  produced.  Fig.  434  shows  the  m.m.fs.  of  the  three 
phases  as  vectors. 

The  currents  are: 

^'i  =  Im  cos  6,  in  phase  1, 

i?  =  Im  cos  (0  —  120),  in  phase  2, 

&3  =  Im  cos  (0  -  240),  in  phase  3. 


INDUCTION  MOTOR  449 

The  m.m.f.  of  phase  1  is  nj.m  cos  6  in  direction  OA. 
The  m.m.f.  of  phase  2  is  nj.m  cos  (6  —  120)  in  direction  OB. 
The  m.m.f.  of  phase  3  is  nilm  cos  (0  —  240)  in  direction  OC. 
The  resultant  m.m.f.  in  the  horizontal  direction  is 

mx  =  njm  cos  B  +  njm  cos  (0  -  120)  cos  120  +  njm  cos  (0  -  240) 
cos  240 

=  njm\  cos  0  —  ~(cos  0  cos  120  +  sin  0  sin  120) 

I  a 

-  »  (cos  0  cos  240  +  sin  0  sin  240) 


=  niTO  »  cos  0  =  2  nim  cos 

The  resultant  m.m.f.  in  the  vertical  direction  is 
mY  =  njm  {cos  (0  -  120)  sin  120  +  cos  (0  -  240)  sin  240} 

=  njm  \  ^r-(cos  0  cos  120  +  sin  0  sin  120)  -^TT  (cos  e  cos  24° 
(    4 

+  sin  0  sin  240) 

Q 

sin  0  =  ^  wt7TO  sin  0. 
The  resultant  m.m.f.  of  the  three  phases  is 


m  =  vW2  +  my2  =  %wil«  V cos2  0  +  sin2  0  =  %nilm 

and  makes  an  angle  0  with  the  OX  axis. 

The  resultant  m.m.f.  is,  therefore,  constant  in  value  being 
equal  to  %  times  the  maximum  m.m.f.  of  one  phase  and  it  re- 
volves at  synchronous  speed. 

This  constant  m.m.f.  produces  a  field  of  constant  strength 
revolving  at  synchronous  speed.  The  revolving  field  links  suc- 
cessively with  the  windings  and  generates  e.m.fs.  in  them.  The 
flux  linking  with  any  phase  is  maximum  when  the  current  in  that 
phase  is  maximum  and,  therefore,  the  flux  and  current  are  in 
phase. 

If  a  four-pole  stator,  Fig.  431,  had  been  chosen  instead  of  the 
two-pole  stator,  the  m.m.fs.  of  the  two-phase  windings  would 
have  been  combined  at  45  degrees  instead  of  90  degrees  and  the 
resultant  m.m.f.  and  flux  would  not  remain  constant  but  would 
pulsate  four  times  during  each  revolution.  The  flux  threading 
any  phase  would,  however,  still  vary  according  to  a  sine  law  and 
would  be  in  phase  with  the  current  in  that  phase, 

29 


450 


ELECTRICAL  ENGINEERING 


To  reverse  the  direction  of  rotation  of  a  two-phase  induction 
motor,  it  is  necessary  to  reverse  one  phase  only. 

To  reverse  a  three-phase  motor  any  two  leads  may  be  inter- 
changed. 

378.  The  Rotor. — The  secondary  or  rotor  is  made  in  two 
forms :  (a)  the  wound  rotor,  and  (6)  the  squirrel-cage  rotor.  The 
wound  rotor  consists  of  a  laminated  iron  core  with  slots  carrying 
the  winding,  which  must  have  the  same  number  of  poles  as  the 
stator  winding  but  may  have  a  different  number  of  phases.  It  is 
usually  wound  for  three  phases  and  the  ends  of  the  windings  are 
brought  out  to  slip  rings  so  that  resistances  may  be  inserted  in  the 
windings  for  starting  and  the  terminals  short  circuited  under 
running  conditions. 


OD 


irjril  I 


FIG.  435. — Squirrel-cage  rotor. 

The  squirrel-cage  rotor  winding  consists  of  a  number  of  heavy 
copper  bars  short-circuited  at  the  two  ends  by  two  heavy  brass 
rings,  Fig.  435.  The  construction  is  very  rugged  and  there  is 
nothing  to  get  out  of  order. 

When  the  rotor  with  its  closed  windings  is  placed  in  the  revolv- 
ing magnetic  field  produced  by  the  stator  currents,  the  flux  cuts 
across  the  conductors  on  the  rotor  and  generates  e.m.fs.  in  them. 
Currents  flow  in  the  rotor  equal  to  the  e.m.fs.  divided  by  the  rotor 
impedances.  These  currents  reacting  on  the  magnetic  field 
produce  torque  and  the  rotor  revolves  in  the  direction  of  the  field. 
At  no  load  the  rotor  runs  almost  as  fast  as  the  field  and  very  small 
e.m.fs.  and  currents  are  induced  in  its  conductors.  When  the 
motor  is  loaded  the  rotor  lags  behind  the  field  in  speed  and  large 
currents  are  induced  to  give  the  required  torque. 

379.  Slip. — The  difference  between  the  synchronous  speed  or 
speed  of  the  stator  field  and  the  speed  of  the  rotor  is  called  the 
slip  and  is  expressed  as  a  per  cent,  of  synchronous  speed. 


INDUCTION  MOTOR  451 

A  six-pole,  60-cycle  motor  has  a  synchronous  speed 


If  the  speed  of  the  rotor  at  full  load  is  1,176  r.p.m.,  the  slip  is 


X  10°  Per  cent-  =  2  per  cent. 
1,JUU 

The  slip  at  full  load  varies  from  2  to  5  per  cent,  in  motors  designed 
for  constant  speed. 

The  rotor  speed  may  be  expressed  as 

S  =  (I  -  s)  N  r.p.m.  (394) 

380.  Magnetomotive  Force  of  the  Rotor.  —  The  frequency  of  the 
e.m.fs.  and  currents  induced  in  the  rotor  windings  at  a  slip  s  is  sf 
if  /  is  the  frequency  of  the  e.m.fs.  impressed  on  the  stator. 

The  polyphase  currents  in  the  rotor  windings  produce  a  re- 

sultant m.m.f.  revolving  relative  to  the  rotor  at  a  speed  -  = 

sN  r.p.m.  But  the  rotor  itself  is  revolving  at  a  speed  S  = 
(1  —  s)  N  r.p.m.  and  therefore  the  rotor  m.m.f.  is  revolving 
at  a  speed  sN  +  (1  —  s)  N  =  N,  that  is,  at  the  same  speed  as 
the  stator  m.m.f.  The  two  m.m.fs.  are  therefore  stationary  rela- 
tive to  each  other  and  they  are  nearly  opposite  in  phase  as  in  the 
transformer. 

381.  Electromotive  Force  and  Flux  Diagram  for  the  Induction 
Motor. 

Let     TI  =  stator  resistance  per  phase. 

LI  =  stator  self  -inductance  per  phase. 

Xi  =  2-jrfLi  =  stator  reactance  per  phase. 

Zi  =  -\Ai2  +  #i2  =  stator  impedance  per  phase. 

r2  =  rotor  resistance  per  phase. 

L2  =  rotor  self-inductance  per  phase. 

.  x2  =  27r/L2  =  rotor  reactance  per  phase  at  standstill. 

sxz  =  27rs/L2  =  rotor  reactance  per  phase  at  slip  s. 

Z'z  =  v^22  +  #22  =  rotor  impedance  per  phase  at  standstill. 


Z2  =  \/7*22  +  «2#22  =  rotor  impedance  per  phase  at  slip  s. 
In  Fig.  436 

Ei  =  the  e.m.f.  impressed  on  one  phase  of  the  stator. 
1  1  =  the  current  in  one  phase  of  the  stator.     /i  =  IM  +  I'- 
IM  =  the  magnetizing  current  in  one  phase  of  the  stator. 
I'  =  the  load  current  in  one  phase  of  the  stator. 


452 


ELECTRICAL  ENGINEERING 

the  current  in  one  phase  of  the  rotor;  72  =  —I',  where 

the  turns  per  phase  per  pair  of  poles  on  the  stator, 
the  turns  per  phase  per  pair  of  poles  on  the  rotor, 
the  flux  per  pole  linking  one  phase  of  the  stator  which 
would  be  produced  by  the  current  Ii  acting  alone. 

the  part  of  $1  which  crosses  the  gap  and  links  with  one 
phase  of  the  rotor.     $ig  =  v&\,  where  vi  is  a  constant. 


FIG.  436. — Currents,  fluxes  and  e.m.fs.  in  an  induction  motor. 

>iL  =  the  part  of  $1  which  does  not  cross  the  gap,  but  is  the 
leakage  flux  of  self -inductance  of  the  stator. 

<i>2  =  the  flux  per  pole  linking  one  phase  of  the  rotor,  which 
would  be  produced  by  the  current  1 2  acting  alone. 

$2    =    $20   +  3>2L- 

>2<7  =  the  part  of  $2  which  crosses  the  gap  and  links  with  one 
phase  of  the  stator.  ^>2ff  =  ^2^2,  where  vz  is  a  constant. 

>2L  =  the  part  of  $2  which  does  not  cross  the  gap,  but  is  the 
leakage  flux  of  self -inductance  of  the  rotor. 

$a  =  the  actual  flux  per  pole  crossing  the  gap  and  linking 
with  one  phase  of  both  stator  and  rotor ;  it  is  the  result- 
ant of  3>ia  and  <i>20. 

$,  =  actual  flux  per  pole  linking  one  phase  of  the  stator  it ;  is 
the  resultant  of  <£„  and  $IL- 


INDUCTION  MOTOR 


453 


E 


E8 


3>r  =  the  actual  flux  per  pole  linking  one  phase  of  the  rotor; 

it  is  the  resultant  of  3>ff  and  $2L. 
Eig  =  the  back  e.m.f.  generated  in  one  phase  of  the  stator  by 

the  flux  3>ff. 
=  the  e.m.f.  of  self-inductance  generated  in  one  phase  of 

the  stator  by  the  leakage  flux  $11,.     EIL  =  Ii%i> 
=  the  back  e.m.f.  generated  in  one  phase  of  the  stator  by 

the  flux  $8.     Es  is  the  resultant  of  Eig  and  EIL. 
EI  =  the  e.m.f.  impressed  on  one  phase  of  the  stator.     It 
must  be  exactly  equal  and  opposite  to  E3  if  the  stator  £e- 
sistance  drop  /  \r  i  is  neglected.     This  drop  is  of  the  order 
of  2  per  cent,  at  full  load. 

Since  EI  is  constant,  Es  and  $s  must  be  constant. 
E2g  =  the  e.m.f.  generated  in  one  phase  of  the  rotor  by  the 

flux  <1>0.     E2a  =  s—Eig  =  sE2.  where  E2  is  the  e.m.f. 
HI 

which  would  be  generated  in  one  phase  of  the  rotor  by 
the  flux  <f>0  at  standstill. 


FIG.  437. 

E2L  =  the  e.m.f.  of  self-inductance  generated  in  one  phase  of 

the  rotor  by  the  leakage  flux  $2L.     E2L  =  sI2x2. 
Er  =  the  e.m.f.  generated  in  one  phase  of  the  rotor  by  the 
flux  <l>r.     Er  is  the  difference  between  E2g  and  E2L  and  is 
equal  to  1 2r2.     It  is  in  phase  with  the  rotor  current  I2 
and  lags  90  degrees  behind  the  flux  3>r. 

As  the  induction  motor  is  loaded  the  end  d  of  the  vector  $1  fol- 
lows a  circle  passing  through  /  and  having  its  center  on  of  pro- 
duced (Fig.  437). 


454 


ELECTRICAL  ENGINEERING 


382.  Proof  that  the  Locus  is  a  Circle. — From  d  draw  dk  at 
right  angles  to  fd  to  cut  of  produced  in  k.  Then  the  semicircle 
fdk  is  the  locus  of  d. 

In  the  triangles  aob  and  fdk 


therefore 
and 


Z  oab  =  Z  fdk,  being  right  angles, 
<  oba  =  Z  dfk, 

fk_fd_       fd  fd 

ob       ab       ac  —  cb       oh  —  cb 


ob  X 


Vi-of  X  v*-  oh 


o/i  —  c6         0/1  — 


_     . 


1  — 


a  constant 


since  of  is  constant. 

Therefore  the  locus  of  d  is  a  circle  described  on  the  diameter  fk. 

383.  Magnetomotive  Force  Diagram.  —  Since  the  magnetic 
circuit  of  the  machine  is  not  saturated  by  the  fluxes  $8  and  <l>r,  the 
flux  diagram,  Fig.  437,  may  be  replaced  by  the  m.m.f  .  diagram, 
Fig.  438. 


total  m.m.f.  of  the  stator  per  phase. 
nil'  =  m.m.f.  of  the  load  component  of  stator  current. 
n-J-M  =  m.m.f.  of  the  magnetizing  current. 
n2J2  =  m.m.f.  of  the  rotor  per  phase  ;  it  is  equal  and  opposite 
to  njf. 


FIG.  438. — M.m.f.  diagram. 


FIG.  439. — Stator  current  diagram. 


384.  Stator   Current   Diagram.  —  The   m.m.f.    diagram,    Fig. 
438,  may  be  replaced  by  the  stator  current  diagram,  Fig.  439. 

Jx  =  total  current  tin  one  phase  of  stator.     I\  =  IM  +  /'• 
IM  =  magnetizing  current  in  one  phase  of  stator. 
I'  =  load  component  of  current  in  one  phase  of  stator. 


72  =  current  in  one  phase  of  rotor.     72 


INDUCTION  MOTOR 


455 


fd  =  /'  =  — J2  =  load  component  of  stator  current  per  phase 

and  represents  the  rotor  current  per  phase. 

To  obtain  an  expression  for  the  diameter  of  the  circle,  consider 
the  special  case  where  the  motor  is  at  standstill  and  the  rotor  and 
stator  resistances  are  both  zero. 


0     / 


FIQ.  440. — Induction  motor  at  standstill. 


Referring  to  Fig.  440 


E1  =  E8  =  Elg 


and 
therefore, 


rl   ^2; 


#1 


-     I'x, 


if  the  magnetizing  current  IM  is  neglected  in  comparison  with  the 
load  component  of  stator  current  /'. 

This  value  of  I'  is  represented  by  fk  and  is  the  diameter  of  the 
circle.     Its  value  is 

-  (395) 


The  diameter  of  the  circle,  therefore,  varies  directly  as  the 
impressed  e.m.f.  and  inversely  as  the  sum  of  the  reactances  and 
therefore  as  the  frequency  of  the  supply. 

Thus  as  the  motor  is  loaded  the  end  d  of  the  vector  representing 
the  stator  current  follows  a  semicircle  of  diameter 

ft 


456 


ELECTRICAL  ENGINEERING 


385.  Rotor  Electromotive   Force   and   Current.  —  The   e.m.f. 
generated  in  one  phase  of  the  rotor  at  slip  s  by  the  flux  3>ff  is 

E2g  =  sK$g,  where  K  is  a  constant, 


=  sEz,  where  Ez  =  K$g  is  the  e.m.f.  which  would 
be  generated  in  the  rotor  at  standstill  by  the  flux  <i>ff.  Ez  does 
not  remain  constant  as  the  motor,  is  loaded,  since  the  flux  <£a  does 
not  remain  constant  but  decreases  about  30  per  cent,  from  no  load 
to  standstill  when  the  rotor  is  locked  and  slip  is  unity.  Up  to 
full  load  &g  and  Ez  may  be  considered  to  remain  constant. 
The  impedance  of  the  rotor  at  slip  s  is 

Z2  -  Vr22  4-  s2z22  (369) 

The  rotor  current  is 


T  Oi^2 

=     ~Z~ 

The  rotor  power  factor  is 

cos  62  = 

E! 


(397) 


(398) 


FIG.  441. 

386.  Rotor  Input. — The  power  transferred  from  the  stator  to 
the  rotor  per  phase  is  the  product  of  the  back  voltage  E',  gener- 
ated in  the  stator  by  the  flux  $g,  the  load  component  of  the 
stator  current  I'  and  the  cosine  of  the  angle  between  them ;  it  is 

pr  =  ET  cos  02  (see  Fig.  441)  (399) 

but  E'  =  E±    =  —E- 

n2 

and  T/       ^2  r  . 


INDUCTION  MOTOR  457 

therefore, 

pr  =  ET  cos  02  =  —  E2  —  h  cos  02  =  #2/2  cos  <92.     (400) 
n2       HI 

Thus  the  power  input  to  the  rotor  per  phase  is  the  product  of  the 
e.m.f.  which  would  be  generated  in  the  rotor  by  the  flux  <$>ff  at 
standstill  and  the  power  component  of  the  rotor  current. 
The  total  power  input  to  the  n  phases  of  the  rotor  is 

Pr  =  nE2I2  cos  62  (401) 

387.  Rotor  Copper  Loss  and  Slip. — The  power  consumed  by 
the  rotor  copper  loss  per  phase  is 

/22r2  =  sE2I2  cos  62,  (402) 

and  for  the  n  phases  it  is 

Lr  =  n/2V2  =  nsE2I2  cos  02.  (403) 

01.  ,.    rotor  copper  loss      Lr       snE2I2cosB2  ,._.. 

Slip  =  the  ratio  -        — P-         -  =  ~  =  — pr^ -~  =  s.   (404) 

rotor  input  PT       nE2I2  cos  02 

388.  Rotor  Output. — The  rotor  output  per  phase  is 

p  =  pr  —  I22r2  =  E2I2  cos  02  —  sE2I2  cos  02 

=  (I  -  s)  #2/2  cos  02  (405) 

and  the  total  rotor  output  is 

p  =  np  =  n(l  -  s)  #2/2  cos  02  watts  (406) 

•  *  (1  "  S\ff 2  C°S  *2  h.p.  (407) 

From  equations  (401)  and  (406)  the  rotor  output  is 

P-    (l-s)Pr=^Pr 

and  it  is  equal  to  the  rotor  input  multiplied  by  the  rotor  speed  in 
per  cent,  of  synchronous  speed. 

389.  Torque. — If  T  is  the  torque  in  Ib.-ft.  the  output  may  be 
expressed  as 

2irST 

~  33,000     P' 
Thus  the  torque  is 

„,  _  n  (1  -  s)  E2I2  cos  02       33,000  _  _  ^  nE^Ii  cos  02 
'      ' 


7.05       =  7.05  lb..ft.  (408) 

N  sync,  speed 


458 


ELECTRICAL  ENGINEERING 


The  torque  of  an  induction  motor  is  usually  expressed  not  in 
pounds-feet  but  in  synchronous  watts,  that  is,  in  terms  of  the 
power  which  would  be  developed  at  synchronous  speed. 

The  torque  in  synchronous  watts  is 


sync,  watts 


=  PX 


N 
S 


(409) 


and  it  is  equal  to  the  power  input  to  the  rotor. 

390.  Rotor  Efficiency. — Neglecting  all  losses  except  the  rotor 
copper  loss  the  rotor  efficiency  is 


., 

77r  =  •=-  100  per  cent. 
Pr 


n(l  —  s)  E2h  cos  02 


cos 


100  per  cent. 


S 


=  (1  —  s)  100  per  cent.  =  -    100  per  cent. 


(410) 


that  is,  the  rotor  efficiency  is  equal  to  the  rotor  speed  in  per  cent, 
of  synchronous  speed  and,  therefore,  the  efficiency  of  an  induction 
motor  is  always  less  than  the  speed  in  per  cent,  of  synchronous 
speed. 

391.  Modification  of  Diagram. — When  an  induction  motor  is 
running  without  load,  a  current  70  flows  in  each  phase  of  the 
stator  which  has  two  components,  Fig.  442,  IM  the  magnetizing 
current  90  degrees  behind  the  impressed  e.m.f.  E\,  and  IP  the 
power  component  in  phase  with  EI. 


Ip^, 

_       0  a  r 

FIG.  442.       FIG.  443. — Circle  diagram  of  an  induction  motor. 


The  product  nEJP  (where  n  is  the  number  of  phases)  is  the 
power  required  to  supply  the  no-load  losses.  These  are  the  iron 
loss  and  a  small  copper  loss  in  the  stator  and  the  friction  and 
windage  losses  of  the  rotor.  The  iron  loss  in  the  rotor  may  be 
neglected  since  the  rotor  frequency  is  low. 


INDUCTION  MOTOR  459 

The  current  required  to  supply  the  stator  losses  has  no  cor- 
responding component  in  the  rotor,  but  the  power  to  overcome 
the  friction  and  windage  losses  must  be  transferred  from  the 
stator  to  the  rotor  and  therefore  requires  a  current  in 
the  rotor. 

As  the  motor  is  loaded  and  slows  down  the  stator  iron  loss  re- 
mains nearly  constant,  the  friction  and  windage  losses  decrease 
and  the  rotor  iron  loss  increases.  At  standstill  the  friction  and 
windage  losses  are  absent  but  the  rotor  iron  loss  is  large  since  the 
rotor  frequency  is  the  same  as  the  stator  frequency.  The  iron 
friction  and  windage  losses  are  therefore  considered  to  remain 
constant  and  the  small  component  of  rotor  current  required  to 
supply  the  friction  and  windage  losses  is  neglected. 

The  diagram,  Fig.  439,  must  therefore  be  changed  to  Fig.  443 
by  the  addition  of  IP  the  power  component  of  the  stator  current 
per  phase  at  no  load. 

The  diameter  of  the  circle  is  raised  through  the  distance  af  = 
IP  and  of  now  represents  not  the  magnetizing  current  IM  but  the 
no-load  current  /o  =  V^M2  +  Zp2- 


If  os  represents  the  stator  current  per  phase  at  standstill  and 
sw  is  its  power  component,  then,  since  there  is  no  output,  the 
power  input  is  consumed  by  the  losses.  Therefore,  input  = 
losses  =  nEiSw;  constant  losses  =  nEJ.P  =  nEiVW  and  copper 
losses  =  nEiSV  =  n(I"irl  +.  722r2). 

The  stator  copper  loss  is  taken  as  nl'2ri  because  the  stator  cop- 
per loss  at  no  load  is  included  in  the  constant  losses.  It  is  there- 
fore assumed  that  /iVi  =  IO*TI  +  I/2n  +  which  is  approximately 
correct  up  to  full  load. 

Divide  sv  at  t  so  that  st  :  tv  =  722r2  :  7'Vi,  then,  stator-load 
copper  loss  =  nEitv  and  rotor  copper  loss  =  nEist. 

Join  ft  and  from  d  any  point  on  the  circle  to  the  left  of  s  draw 
dmpqr  perpendicular  to  the  diameter  fk.  It  is  to  be  shown 
that  mp  is  the  stator  current  required  to  supply  the  rotor  copper 
loss  for  a  rotor  current  represented  by  fd  and  that  pq  is  the  cur- 
rent to  supply  the  corresponding  stator  load  copper  loss. 

The  rotor  copper  loss  is 

n/2V2  =  n(—  I'Yri  =  KI'2  =  Kfd    =  K(fa   +  qd) 

W2     ' 

=  K  X  fq(fq  +  qk)  =  K  X  fq  X  D  =  fq  X  a  constant, 
since  K  and  D  are  both  constants. 


460  ELECTRICAL  ENGINEERING 

Therefore  the  rotor  copper  loss  is  proportional  to  fq ;  but 

mp  _  st         mp          st 


and  since  st  represents  the  rotor  copper  loss  for  a  current  fs,  mp 
represents  the  rotor  copper  loss  for  a  current  fd. 

Similarly  pq  represents  the  stator  copper  loss  for  stator  cur- 
rent fd. 

392.  Interpretation  of  Diagram. — At  any  value  of  stator  cur- 
rent od  =  /i,  Fig.  443. 

nEi  dr  =  stator  input  in  watts, 
nEi  qr  =  constant  losses, 
nEipq  =  stator  copper  loss, 
nEi  mp  =  rotor  copper  loss, 
nEi  dm  =  rotor  output  in  watts  =  mechanical  load, 

dr  , 

— -,  =  power  factor, 

dm        _  . 

-j-  =  efficiency, 

mp  _  rotor  copper  loss  _     ,. 

dp  rotor  input 

dm  _  rotor  output  _        actual  speed         _  J?  __  -i 

dp        rotor  input        synchronous  speed       N 

The  torque  corresponding  to  output  nEi  dm  is 

-,       nEidm          33,000      „          ,  », 
T  =      „.„     X  n    ,   -    —^  lb.  at  1  ft.  radius. 
746          2ir  (r.p.m.) 

At  synchronous  speed  this  torque  would  represent  an  output 

N  dr) 

nEi  dm  X  TT  =  nEi  dm  X  3*-  =  nE\  dp  watts  =  rotor  input, 
o  dm 

The  torque  in  synchronous  watts  is  equal  to  the  watts  input  to 
the  rotor  =  nEidp. 

At  standstill  the  torque  in  synchronous  watts  is  nEiSt  and 
represents  the  starting  torque  of  the  motor. 

The  maximum  value  of  torque  in  synchronous  watts  is  nEi  X 
maximum  value  of  dp. 

The  maximum  output  in  watts  is  nE\  X  maximum  value  of  dm. 
For  average  25-cycle  motors,  starting  torque  is  one  and  one-half 
to  two  and  one-half  times  full-load  torque ;  starting  current  is  six 
to  eight  times  full-load  current ;  and  maximum  running  torque  is 


INDUCTION  MOTOR  461 

two  and  one-half  to  three  and  one-half  times  full-load  torque. 
For  60-cycle  motors,  starting  torque  is  one  to  one  and  one-half 
times  full-load  torque  ;  starting  current  is  five  to  six  times  full-load 
current  ;  and  maximum  running  torque  is.  two  to  two  and  one-half 
times  full-load  torque. 

393.  Construction  of  Diagram  from  Test  for  a  Three-phase 
Motor.  —  1.  Run  the  motor  light  at  rated  voltage  and  rated  fre- 
quency. Read  impressed  voltage,  current  and  watts  input 
Ei,  IQ  and  WQ. 

J/o  =  of  on  the  diagram. 

0 


V  3E 


=  IP  =  af  =  wv  on  the  diagram. 


2.  Lock  the  rotor  and  impress  reduced  voltage  and  raise  it  until 
twice  full-load  current  flows  in  the  stator.  Read  impressed  vol- 
tage, current,  watts  input  and  torque,  EL,  IL,  WL  and  TL.  To 
get  the  value  of  locked  current  at  rated  voltage,  raise  the  values 
of  IL  in  the  ratio  EI  :  EL.  To  get  the  values  of  locked  watts  and 
locked  torque  at  rated  voltage,  raise  the  values  of  WL  and  TL  in 
the  ratio  Ei2  :  EL2.  To  get  accurate  results  for  the  circle  diagram 
it  is  better  to  reduce  the  values  of  watts  and  torque  to  terms 
of  power  current  per  phase.  This  is  done  by  dividing  the  values 
of  WL  and  TL  by  \/3EL. 

Plot  on  a  base  of  impressed  voltage, 


These  three  loci  should  be  straight  lines  passing  through  the 
origin  and  can  be  produced  until  they  cut  the  ordinate  at  the 
rated  voltage  of  the  motor. 

The  following  results  are  obtained. 

Value  of  IL  at  rated  voltage  =  os  on  the  diagram. 

WL 
Value  of      ._      at  rated  voltage  =  sw  on  the  diagram. 


3.  Measure  the  resistance  of  the  stator  per  phase  =  r\.  Then 
the  stator  copper  loss  locked  at  rated  voltage  is  3  os2  n  watts 

and  -  ,-      =  vt  on  the  diagram. 

The  rotor  copper  loss,  locked  at  rated  voltage,  is  \/3Ei  X  st 
and  is  known  since  st  =  sw  —  wv  —  vt. 


462 


ELECTRICAL  ENGINEERING 


The  rotor  copper  loss  also  represents  the  starting  torque  in 
synchronous  watts  and  therefore  if  T\  is  the  value  of  TL  at  rated 
voltage 

Ti  X  2ir  X  (sync,  speed  r.p.m.)  \/3#i  X  s£ 

33,000  -  Sh°uld  equal  ~  " 


The  circle  diagram  can  be  drawn  in  from  the  values  obtained 
above. 

The  motor  has  been  assumed  to  be  F-connected  and  the  vol- 
tage EI  is  the  line  voltage. 

394.  Test  of  an  Induction  Motor.—  The  following  readings 
were  obtained  from  tests  on  a  30-hp.,  440-volt,  60-cycle,  three- 
phase  star-connected  induction  motor,  having  eight  poles  and  a 
synchronous  speed  of  900  r.p.m. 

Resistance  of  stator  between  terminals  at  normal  operating 
temperature  =  0.50  ohms. 

Resistance  of  stator  per  phase  =  r\  =  0.25  ohms. 


RUNNING   SATURATION   OR   NO-LOAD   TEST 


Impressed 
e.m.f.  =  E 

Exciting  current  = 
Jo 

Watts  input  =  Wo 

550 

23.0 

2,920 

480 

15.8 

1,750 

440 

14.0 

1,475 

405 

12.5 

1,350 

380 

10.9 

1,250 

290 

7.8 

890 

COCKED  TEST 


Impressed 
e.m.f.  =>  EL 

Stator 
current  =  IL 

Watts  input  ' 
=  WL 

Starting 
torque  =  TL 
Ib.-ft. 

.  WL 

TL 

VSEL 

VZEL 

307.5 

121.0 

32,100 

145 

60.3 

0.271 

276.5 

108.5 

25,400 

120 

53.2 

0.251 

257.5 

100.0 

21,000 

99 

47.3 

0.244 

233.5 

88.0 

16,500 

84 

40.9 

0.207 

201.5 

74.0 

11,550 

60 

33.2 

0.170 

164.0 

59.0 

7,350 

35 

25.9 

0.124 

440.0 

170.0 

61,600 

297.5 

81.0 

0.390 

INDUCTION  MOTOR 


463 


These  values  are  plotted  in  Fig.  444  against  impressed  e.m.f. 
and  the  values  at  440  volts  are  obtained  by  producing  the  curves. 

W"  T7 

The  loci  of  IL,    /OET  and     .--,  are  straight  lines  passing  through 


the  origin. 

With  line  voltage  E  =  440,  the  impressed  voltage  per  phase  = 

£:  =  JU^  =  254. 


V3 


Amperes  Exciting  Current  =  I  o 

5          10         15         20         25         30         35        40 


UJ 


400 

I 

£300 

•200 
3 
100 


X 


500  1000      •  1500 

Watts  Input  at  No-load  =  W0 


81     100 


MO      170    Amperes 


600 
440 


50    100   150   200   250   300   350   390 

FIG.  444. 

At  No  Load. 

Stator   current   or  exciting  current  =  70  =  14.0  amp.  =  of 

(Fig.  445). 

Watts  input  =  Wo  =  1,475  =  iron  loss  -f  friction  and  wind- 
age losses  and  these  are  assumed  to  be  constant. 

Wo  1,475 


Power  factor  =  cos  00 


=  0.137 


V3EIQ       V3  X  440  X  14 
=  13.7  per  cent. 
In-phase  current  =  IP  =  I0  cos  00  =  1.92  amp.  =  fa  =  vw. 

Magnetizing  current  =  IM  =  V/o2  —  Ip2  =    \  142  —  1.922  = 
13.8  amp.  =  oa. 


464 


ELECTRICAL  ENGINEERING 


Rotor  Locked. 

Stator  current  =  IL  =  170  amp.  =  os. 
Watts  input  =  WL  =  61,600  =  constant  losses  +  stator  cop- 
per loss  +  rotor  copper  loss. 

WL  61,600 


Power  factor  =  cos  0L 


=  0.476 


V3EIL      V3  X  440  X  170 
=  47.6  per  cent. 
In-phase  component  of  stator  current  =  IL  cos  8L  =  81  amp. 


=  sw. 


StanditlU 


E,=. 


FIG.  445. 

Quadrature  component  of  stator  current  =  IL  sin  BL  =  149 

amp.  =  ow. 
Stator  copper  loss  =  3/Vi  =  3  X  1702  X  0.25  =  21,700 

21  700 
watts,  represented  by  the  current  tv  =  — 7=^ — :  -  =  28.4 


V3  X440 


amp. 


Rotor  copper  loss  =  61,600  -  1,475  -  21,700  =  38,425,  rep- 

38,425 


resented  by  the  current  st  = 


\/3  X440 


=  50.5  amp. 


The  points  o,  /,  a,  s,  t,  v  and  w  are  therefore  fixed  and  the  circle 
may  be  drawn  in,  as  shown  in  Fig.  445. 

The  starting  torque  at  440  volts  =  0.39  X  V3  X  440  =  297.5 
Ib.-ft.  and  the  corresponding  output  at  synchronous  speed  = 
2  X  3.14  X  900  X  297.5 


=  51 


'  =  51  X 


33,000 

746  =  38,046  synchronous  watts  =  rotor  input  =  rotor  copper 
loss  and  should  agree  with  the  value  38,425  above.  This  serves 
as  a  check  on  the  circle. 


INDUCTION  MOTOR  465 

For  full-load  output  of  30  hp.  the  vertical  intercept  between 
the  circle  and  the  line  fs  must  be  the  required  in-phase  current 

30  X  746 

=  —7= -  =  29.3  amp.  =  dm.    /The  full-load  point  d  is  thus 

V3  X  440  I 

fixed  and  the  corresponding  stator  current  =  l\  =  od  =  40  amp., 

power  factor  =  cos  0i  =  --,  =  0.8625  =  86.25    per    cent.,  effi- 
•j  od  j 

ciency  rj  =  -?—  100  per  cent.  =  84.3  per  cent.,  speed  =  S  =  -7—  X 
sync,  speed  =  0.9275  X  900  =  835  r.p.m.,  and  the  slip  =  s  = 

^  =  0.0725  =  7.25  per  cent. 
dp 

At  one-half  load,  I\  —  23  amp.,  cos  0i  =  75  per  cent.,  17  =  84.0 
per  cent.,  S  —  870  r.p.m.,  s  =  3.3  per  cent.;  at  one  and  one-half 
load,  7i  =  65  amp.,  cos  0i  =  85.5  per  cent.,  77  =  77.5  per  cent., 
S  =  779  r.p.m.,  s  =  13.4  per  cent. 

Maximum  torque  or  pull-out  torque  is  represented  by  the  maxi- 
mum intercept  between  the  circle  and  the  line/£  =  73.6  amp.  = 
73.6  X  \/3  X  440  =~5fi,000  "sync. ;  watts~^l407o  Ib.-ft.  The 
slip  for  maximum  torque  is  36.4  per  cent. 

Maximum  output  or  stalling  load  is  represented  by  the  maxi- 
mum intercept  between  the  circle  and  the  line/s  =  57.6  amp.  = 
57.6  X  V3  X  440  =  44,000  watts  =  58.9  hp.  and  the  corre- 
sponding slip  is  24.4  per  cent. 

395.  Analysis  by  Rectangular  Coordinates. — Using  the  method 
of  rectangular  coordinates  called  the  symbolic  method  the  per- 
formance characteristics  of  an  induction  motor  can  be  determined 
if  the  constants,  of  the  motor  are  known. 
Let 

Y  =  9  ~  fo  =  stator  exciting  admittance  per  phase  at  rated 

voltage. 

Z\  =  7*1  +  jxi  =  stator  impedance  per  phase. 
?z  =  r2  +  jsx2  =  rotor  impedance  per  phase  at  slip  s. 
Assume  that  the  ratio  of  turns  is  HI  :  n2  =  1  :  1  and  take  as  real 
axis  of  coordinates  the  e.m.f.  generated  in  the  stator  by  the  flux 
of  mutual  inductance.     The  quantities  used  refer  to  one  phase 
of  the  stator  and  the  corresponding  phase  of  the  rotor. 
E'  =  e.m.f.  generated  in  the  stator. 
Ez  =  Ef  =  e.m.f.  generated  in  the  rotor  at  standstill. 
sE2  =  e.m.f.  generated  in  the  rotor  at  slip  s. 
These  three  e.m.fs.  lie  along  the  real  axis. 

30 


466  ELECTRICAL  ENGINEERING 

The  following  equations  show  the  relations  between  the  various 
e.m.fs.  and  currents  at  slip  s. 
Rotor  current: 


where 


Stator  load  current  : 

/'  =  j2  =  E'(al-ja2).  (413) 

Stator  exciting  current: 

70  =  E'Y  =  E'  (g  -  jb.)  (414) 

Total  stator  current: 

/i  =  /'+/„  =  E'{  (a,  +  g)-j  (a,  +  b)  }  =  E'(b,  -  jb2)          (415) 
where  61  =  0,1  +  Q  and  62  =  a2  +  &  (416) 

E.m.f.  impressed  on  the  stator: 

#i  =  J£'  +  /!?!  =  E'  +  E'  (b,  -  J62)(n  +  jxi) 

=  #'{(1  +  bin  +  6ja?i)  +  j  (6ix,  -  62rO  =  E'  (Cl  +  jc2) 

(417) 

where  Ci  =  1  +  6iri  +  b2Xi  and  c2  =  61X1  —  bzri  (418) 

The  constant  absolute  value  of  the  impressed  e.m.f.  is 

El  =  E'  Vc7T~^.  (419) 

Thus  the  e.m.f.  generated  in  the  stator  is 

W  =  —  pJL=.  (420) 

Vci2  +  c22 

Substituting  this  value  for  E'  in  the  equations  above  the  abso- 
lute values  of  the  various  quantities  as  slip  s  can  be  obtained. 
Rotor  current: 


/2  =  /'  =  E'       aS  +  a22  =  E,  -  (421) 

Vci2  +  c22 
Exciting  current: 

(422) 


+  c2 
Stator  current: 


-.  (423) 


INDUCTION  MOTOR  467 

The  torque  in  synchronous  watts,  was  found  in  Art.  389  to  be 
equal  to  the  rotor  input  in  watts,  which  is 

Pr  =  nE2h  cos  02  =  n#'X#'ai  =  n#'2ai  =  n#i2     2  ^     2  (424) 

Ci    T  £2 

where  n  is  the  number  of  phases. 
The  torque  in  pounds-feet  is  from  Art.  389. 

r  =  7.05£  =  7-05^^-,  (425) 

N  N        ci2  +  c22 

The  rotor  output  is 

p  =  <!-«)  F,-  *tf1?2i|L_£  (426) 

The  stator  power  factor  is  cos  0i,  where  0i  is  the  angle  of  lag 
of  the  current  Ii  behind  the  impressed  e.m.f.  E\.    61  =  0'  +  6", 

0'  =  cos-1      .   Cl  —  -  =  the  angle  by  which  Ei  leads  E'  and 

V  ci2  +  c22 

"  1          =  the    angle   of    lag  of    Ii    behind  E'  '; 


therefore 

cos  0i  =  cos  (0'  +  0")  =  cos  0'  cos  0"  -  sin  0'  sin  6" 

(427) 


V+&22 
The  power  input  to  the  stator  is 


P,  =  nB171  cos  9l  =  ^  X  ^  X 


C22 


(428) 


Using  these  equations  the  various  quantities  can  be  calculated 
in  terms  of  the  slip  and  the  characteristic  curves  of  the  motor 
plotted. 

To  obtain  expressions  for  the  maximum  torque  and  output 
equations  (425)  and  (426)  may  be  differentiated  with  respect  to 
s  but  the  work  is  very  complicated.  Much  simpler  expressions, 
which  are  quite  accurate  enough,  may  be  obtained  by  neglecting 
the  exciting  current  and  deriving  new  equations  for  torque  and 
output  on  this  basis,  as  follows: 

Stator  load  current  : 


468  ELECTRICAL  ENGINEERING 

Stator  impressed  e.m.f . : 


and 


a*)         (429) 


Xl  +  X-L 


and 


Power  input  to  the  stator  is 


(430) 


(431) 


Pi  =  nEi  X  El 


+?) 


(ri  +  - 

V  o 

Power  input  to  the  rotor  is 
Pr  =  Pi  —  stator  copper  loss  =  PI  —  tt//2n 


(432) 


(433) 


and  this  is  equal  to  the  torque  in  synchronous  watts. 
Torque  in  pounds-feet  is 


7.05 


(434) 


INDUCTION  MOTOR  469 

Rotor  output  is 

nES-(l  -  s) 
P  =  (1  -  s)Pr  = — 4 -'  (435) 


To  obtain  an  expression  for  the  maximum  torque,  differentiate 
(434)  with  respect  to  s: 

2L  -  7-05    E  2    A  I  l  \ 

ds          N        l    2  ds  I 


=  0 

for 
max. 


Therefore 


and  from  this  the  slip  at  maximum  torque  is 


_ 

.  torque   - 


/  -  -  — 

V  rt2  + 


(436) 


Substituting  this  value  of  s  in  (434)  the  maximum  torque  is 
found 


7.05  nEi2\/ri2  +  (xi  +  x2)2  

£1  +  Xz)2 

(437) 


N    (n  +  Vri2  +  (xi  +  ^2)2)2  +  (xi  + 
7.05  nEJ 


N    2[n  +  Vn2  +  (xi  +  a;2)2] 
Stator  current  for  maximum  torque  is  found  by  substituting  the 
value  of  slip  in  436  in  equation  (431) : 

•*max.  torque    ==         .  \*do) 

and  it  is  independent  of  the  rotor  resistance  r2. 

Starting  torque  may  be  found  by  substituting  s  =  1  in  (434) : 
7.05  n#!2r2  7.05    f    2   ,      (      . 

'.   =  ~^T  TT  ^o     .     /  -^-9   =  ~^T-  n(I*  rV        (439) 


The  starting  current  is  found  by  substituting  s  =  1  in  (431): 

/starting   =   —p  ®^  '  (440) 

V(ri  +  r2)2  +  (X!  +  x2)2 


470  ELECTRICAL  ENGINEERING 

The  rotor  resistance  corresponding  to  maximum  starting  torque 
is  found  by  substituting  s  =  1  in  (436).     It  is: 


^2  (for  max.  starting  torque)    =    "\/Ti2  +    (#1  +  Xz)2>  (441) 

The  slip  corresponding  to  maximum  output  may  be  found  by 
differentiating  P  with  respect  to  s  and  equating  to  zero  ;  the  oper- 
ation is  rather  long  and  the  result  only  will  be  given  here  : 

Smax.   output   =  /  ==  ==^'  (442) 

TY+  V  (ri  +  r2)2  +  (xi  +  z2)2 

Substituting  this  value  of  s  in  equation  (435)  the  maximum 
output  or  stalling  load  is  found  : 


--  =-.     (443) 

ra)  +  V(ri  +  ra)»+(si  +  *2)«} 

These  equations  are  quite  accurate  except  when  the  exciting 
current  is  very  large.  When  calculating  the  stator  power  factor 
and  the  stator  current  up  to  full  load  the  exciting  current  cannot 
be  neglected. 

From  equations  (437)  and  (443)  it  is  seen  that  both  the  maxi- 
mum torque  and  the  maximum  output  are  proportional  to  the 
square  of  the  impressed  e.m.f.  and  therefore  any  decrease  in  the 
line  e.m.f.  will  seriously  affect  the  motor  characteristics. 

396.  Characteristics  of  an  Induction  Motor  by  the  Symbolic 
Method.  —  Before  applying  this  method  the  constants  g,  b,  TI,  rz, 
Xi  and  #2  must  be  determined.  Taking  as  an  example  the  motor 
of  Art.  394,  the  constants  may  be  found  as  follows: 

440 
Impressed  e.m.f.  per  phase  =  E\  —  —  -7=  =  254  volts. 

v  3 

Stator  resistance  per  phase  =  rx  =  0.25  ohms. 
Stator  current  per  phase  at  no  load  =  70  =  14  amp. 
Watts  input  at  no  load  =  W0  =  1,475. 

Power  factor  at  no  load  =  cos  00  =  —7=-  ~  =  -7— 

V3#/o       V  3  X  440  X  14 

=  0.137  =  13.7  per  cent. 
Stator  exciting  admittance  per  phase  =  Y  =  ^-  =  oc7  =  55.2 

X  10-3. 

Stator  exciting  conductance  per  phase  =  g  =  Y  cos  60  =  7.6 
X  10-3. 

Stator  exciting  susceptance  per  phase  =  b  =  V~Y2  —  g2  = 
54.0  X  10~3. 


INDUCTION  MOTOR  471 

Stator  current  per  phase  locked  =  IL  =  170  amp. 
Watts  input  locked  =  WL  =  61,600. 

i     i    A  WL  61,600 

Power  factor  locked  -  COB  fc  -^-  =  V§  x  44Q  x  17Q  = 

0.476  =  47.6  per  cent. 

T7» 

Equivalent  impedance  per  phase  locked  =  ZL  =  -y-1  = 

1.49  ohms. 

Equivalent  resistance  per  phase  locked  =  RL  =  ZL  cos  6L  = 
0.71  ohms. 

Equivalent  reactance  per  phase  locked  =  XL  =  \/ZL2  —  RL2  = 
1.31  ohms. 

Rotor  resistance  per  phase  referred  to  the  stator  =  r2  =  RL  — 
ri  =  0.71  -  0.25  =  0.46  ohms. 

Assuming  that  the  rotor  reactance  per  phase  referred  to  the 
stator  is  equal  to  the  stator  reactance  per  phase  : 

Xi  =  x2  =  -^  -  =  ~V~  =  0.655  ohms. 

Z  Jj 

The  characteristic  curves  may  be  obtained  by  substituting  any 
values  of  slip  in  the  equations  in  Art.  395  using  the  constants 
found  above. 

It  will  be  sufficient  to  check  the  results  of  this  method  at  one 
or  two  values  of  slip  against  the  results  obtained  from  the  circle 
diagram. 

When 

s  =  0.05: 


01  =  -Tr-a    I*  =  °-095>  «*  =  ~2-n          2  =  0.00764, 

r22  +  s2xz2  rz2  +  s2x22 

bi  =  ai  +  g  =  0.1026,  62  =  «2  +  b  =  0.0626, 

Cl  =  1  +  &iri  +  62^i  =  1.066,       c2  =  biX!  -  b2ri  =  0.0514. 

ft     2    J_    7»   2 

Stator  current  =  /i  =  EiJ~~  —  \  =  28.6  amp. 

\  Ci       i     C2 

The  check  values  from  the  circle  diagram  will  be  taken  at  this 
value  of  stator  current. 

Power  factor  =  cos  Bl  =  c^  ~  ^2  =  Q  g25  = 


82.5  per  cent,  (circle  gives  81.8  per  cent.) 
Output  =  P  =  nE^l~^a^  =  15,300  watts  =  20.5  hp. 
(the  circle  diagram  gives  15,225  watts  =  20.4  hp.). 


472  ELECTRICAL  ENGINEERING 


J,  -  a842  '  84'2  per 

cent,  (circle  gives  85  per  cent.) 
Torque  .r.Z^*!.-^_.  126.5  Ib.-ft. 

(circle  gives  125.5). 

Slip  at  28.6  amp.  from  the  circle  is  4.55  per  cent. 

When  s  =  0.10: 

ai  =  0.212,  «2  =  0.0302,  bi  =  0.22,  62  =  0.085,  ci  =  1.11, 
c2  =  0.123.  7i  =  53.4  amp.,  cos  61  =  88.4  per  cent.,  P  = 
29,250  watts,  77  =  81.3  per  cent.,  T  =  255  Ib.-ft. 

The  values  obtained  from  the  circle  diagram  for  a  stator  cur- 
rent 1  1  =  53.4  are  cos  0i  =  88.1  per  cent.,  P  =  29,000  watts, 
17  =  80.9  per  cent.,  T  =  252  Ib.-ft.,  s  =0.10. 

At  standstill  when  s  =  1: 

oi  =  0.716,  «2  =  1.02,  61  =  0.724,  62  =  1.075,  Cl  =  1.885, 
c2  =  -  0.2185,  /i  =  168  amp.  and  T  =  300  Ib.-ft.;  the  cor- 
responding values  from  the  circle  diagram  are  I\  =  170  and 
T  =  297.5  Ib.-ft. 

Slip  for  maximum  torque,  equation  (431),  is 

Smax.  torque  =  —  /  =  0.345  =  34.5  per  cent. 

Vri2  +  (xi  +  Z2)2 

Maximum  torque,  equation  (437),  is 

= 


max. 

N    2[n 

From  the  circle  diagram,  !Fmax.  =  440  Ib.-ft.  and  the  corre- 
sponding slip  =  36.4  per  cent. 

Slip  for  maximum  output,  equation  (442),  is 

s  max.  output  =  -  ,-  —  =====  =  23.6  per  cent. 


Maximum  output,  equation  (443),  is 

Pmax.   =  —r J1     l      =  =;   =  44,000  watts 

=  58.9  hp. 

From  the  circle  diagram,  P  max  =  44,000  watts  =  58.9  hp.  and 
the  corresponding  slip  is  24.4  per  cent. 

From  the  calculations  above  it  is  seen  that  the  two  methods 
check  very  satisfactorily. 


INDUCTION  MOTOR 


473 


397.  Methods  of  Starting. — Except  in  the  case  of  small  ma- 
chines, induction  motors  should  not  be  started  by  connecting 
them  directly  to  the  mains,  since  the  large  starting  current  at 
low  power  factor  disturbs  the  voltage  regulation  of  the  system. 

Two  methods  of  reducing  the  starting  current  are  in  use.  (1) 
The  voltage  impressed  on  the  stator  is  reduced  by  using  an  in- 
duction starter  which  is  simply  an  auto-transformer  with  one  or 
more  taps  (Fig.  374).  (2)  Resistance  is  inserted  in  series  with 
the  rotor  windings. 

1.  When  the  impressed  voltage  is  reduced,  the  starting  current 
is  reduced  in  proportion  to  it,  but  the  starting  torque  is  reduced 
as  the  square  of  the  voltage. 


FIG.  446. — Starting  on  reduced          FIG.  447. — Starting  torque  with  various  rotor 
voltage.  resistances. 

In  Fig.  446  E\.s\t\  represents  the  starting  torque  of  the  motor 

at  full  voltage  and  osi  represents  the  starting  current,  neglecting 

pi 

the  exciting  current,  and  -^-  •  s2£2  represents  the  starting  torque  at 

& 

half  voltage  and  os2  represents  the  starting  current. 

Since  os2  =  ~-£  the  starting  current  is  reduced  to  one-half  its 

value  at  full  voltage,  but  the  starting  torque  is  reduced  to  one- 
quarter.  The  power  factor  is  not  changed. 

Thus  starting  with  reduced  voltage  gives  very  small  starting 
torque  and  low  power  factor. 

A  squirrel-cage  rotor  may  be  used. 

2.  When  resistance  is  inserted  in  the  rotor  windings  the  start- 
ing current  is  reduced  and  is  brought  more  nearly  in  phase  and 
the  starting  torque  is  increased.  • 

In  Fig.  447  s02o  represents  the  starting  torque  wheD  the  rotor 
circuits  are  closed  without  any  starting  resistance. 

s  it  i  is  the  starting  torque  when  resistance  Ri  is  inserted. 


474 


ELECTRICAL  ENGINEERING 


s2tz  is  the  starting  torque  when  resistance  R2>Ri  is  inserted. 
s3Z3  is  the  maximum  possible  starting  torque  and  is  obtained 
by  inserting  a  resistance  RS>R2;  it  is  the  same  as  the  maximum 
running  torque  of  the  motor.  Rs  -}-  r%  =  A/ri2  +  (#1  +  ^2)2> 
equation  (441).  os0,  osi,  os2  and  os3  are  the  corresponding  stator 
currents  and  cos  0o,  cos  0i,  etc.,  are  the  power  factors  at  start. 

The  curves  in  Fig.  448  are  the  "speed-torque"  characteristics 
for  the  motor  operating  with  the  various  resistances  in  the  rotor. 
The  maximum  torque  is  the  same  in  all  cases  but  it  is  reached  at 

different  speeds.     One  current 
curve  holds  in  all  cases. 

If  a  resistance  R 4  >  Rz  is 
inserted  in  the  rotor  windings 
the  starting  current  is  further 
reduced  and  the  power  factor 
is  improved  but  the  starting 
torque  is  decreased.  R 4  may 
be  made  of  such  value  that 
the  starting  torque  s4£4  is 


equal  to  full-load  torque  and 
the  starting  current  oSi  is 
equal  to  full-load  current. 
Curve  (4)  is  the  speed-torque 
characteristic  for  this  case. 

Thus  by  inserting  resistance 

in  the  rotor  any  starting  torque  up  to  the  maximum  running 
torque  or  "pull-out"  torque  may  be  obtained.  The  starting 
current  is  reduced  and  the  power  factor  is  improved. 

In  starting  a  heavy  load  resistance  Rs  is  used  and  the  motor 
gives  its  maximum  torque  at  start.  The  resistance  is  then  cut 
out  gradually  as  the  speed  increases  and  the  motor  operates  with 
short-circuited  rotor  with  characteristics  as  shown  in  curve  (0). 

If  the  load  to  be  started  is  not  very  great  and  a  large  starting 
current  at  low  power  factor  is  objectionable,  resistance  R*  is  used 
and  the  motor  starts  with  full-load  torque  and  draws  full-load 
current. 

This  second  method  of  starting  requires  a  wound  rotor  with 
slip  rings  and  large  starting  resistances  which  is  much  more 
expensive  than  a  squirrel-cage  rotor. 

For  the  same  line  current,  resistance  starting  gives  about  four 
times  the  torque  given  at  reduced  voltage. 


0      25      50      75    100    125  150    175    200   225    250 
Torque  in  Percent  of  Full  Load 

FIG.  448. — Speed-torque  character- 
istics of  an  induction  motor  with  various 
rotor  resistances. 


INDUCTION  MOTOR 


475 


398.  Applications. — The  constant-speed  or  squirrel-cage  in- 
duction motor  takes  the  place  of  the  direct-current  shunt  motor 
and  has  very  similar  characteristics.  It  is  of  much  more  simple 
and  rugged  construction  than  the  shunt  motor  and  the  wear 
and  danger  due  to  sparking  are  entirely  eliminated. 

It  should  be  used  where  fairly  constant  power  is  required  for 
long  periods,  where  good  speed  regulation  is  required,  where 
starting  is  infrequent  and  only  average  starting  torque  is  neces- 
sary, where  the  motor  is  exposed  to  dust  or  to  inflammable 
materials  or  is  not  easily  inspected.  It  is  suitable  for  driving  line 
shafting,  for  high-  and  low-speed  centrifugal  pumps,  blowers,  fans, 
etc.  It  must  be  started  on  reduced  voltage  except  for  the  small- 
est sizes. 

The  variable-speed  induction  motor  has  a  wound  rotor  with  its 
terminals  connected  to  slip  rings  so  that  resistance  may  be  intro- 
duced to  vary  the  speed  or  to  give  a  large  starting  torque. 


100 

90 


I  80 

s70 

o  60 


700 
GOO 


•§  50  *'  500 
3       p- 

£40  a  400 

|  30 1  300 

W  20  m  200 


10 


100 
0 


50 


100 


250 


350 


150         200 
Torque  Ft.  Lb» 

FIG.  449. — Characteristic  curves  of  a  three-phase,  60-cycle,  220-volt,  20  horse- 
power induction  motor. 

It  should  be  used  where  frequent  starts  under  load  are  neces- 
sary, or  where  the  motor  is  large  enough  to  have  a  bad  effect  on 
the  regulation  of  the  system  due  to  the  large  starting  current  at 
low  power  factor,  as  for  cranes,  elevators,  hoists,  etc. 

The  squirrel-cage  motor  with  a  comparatively  high-resistance 
rotor  can  be  used  where  fairly  large  starting  torque  is  required 
and  where  a  wound-rotor  motor  is  not  advisable,  as  in  cement 
mills,  etc. 

Fig.  449  shows  the  characteristic  curves  of  a  three-phase,  60- 
cycle,  220-volt,  20-hp.  induction  motor  with  a  squirrel-cage  rotor. 


476  ELECTRICAL  ENGINEERING 

399.  Speed    Control   of   Induction   Motors. — The   induction 
motor  is  inherently  a  constant-speed  motor  like  the  direct-current 
shunt  motor  but  its  speed  may  be  varied  in  four  principal  ways: 
(1)  By  variation  of  the  impressed  voltage,  (2)  by  inserting  resist- 
ance in  the  rotor  windings,  (3)  by  changing  the  number  of  poles 
and  (4)  by  cascade  control  or  concatenation. 

400.  Voltage  Control. — Since  the  torque  for  a  given  slip  is 
proportional  to  the  square  of  the  impressed  voltage,  the  speed 
may  be  varied  through  a  small  range  by  variation  of  the  voltage. 
This  may  be  accomplished  (a)  by  introducing  resistance  in  series 
with  the  stator  windings  or  (b)  by  using  a  variable  ratio  trans- 
former or  compensator. 

Control  by  series  resistance  is  very  simple  but  is  inefficient 
while  control  by  a  compensator  is  more  expensive  but  gives 
higher  efficiency.  Both  methods  to  be  effective  require  a  motor 
having  a  high-resistance  rotor  with  its  resulting  large  slip.  A 
squirrel-cage  rotor  may  be  used. 

401.  Rotor  Resistance  Control. — The  speed  of  an  induction 
motor  may  be  varied  by  using  a  wound  rotor  with  slip  rings  and 
connecting  external  resistances  into  the  circuit.     With  suitable 
resistances  any  required  speed-torque  curve  may  be  obtained 
(Art.   397).     This  method  of  control  corresponds  to  the  use 
of  resistance  in  series  with  the  armature  of   a  direct-current 
shunt    motor.       It  is  very  inefficient  and  the  speed  changes 
with  load. 

402.  Pole-changing. — By  the  use  of  special  windings  the  num- 
ber of  poles  on  an  induction  motor  may  be  changed  and  therefore 
also  its  synchronous  speed.     Two  or  at  most  three  synchronous 
speeds  may  be  obtained  by  this  means.     Take  for  example  an 
eight-pole  motor  with  a  double-layer  winding  of  two-thirds  coil 
pitch;  its  synchronous  speed  may  be  doubled  by  reconnecting 
for  four  poles  with  a  coil  pitch  of  one  and  one-third.     A  squirrel- 
cage  rotor  is  generally  used  but  if  intermediate  speeds  are  re- 
quired a  wound  rotor  must  be  used  and  its  poles  changed  at  the 
same  time  as  those  on  the  stator.     Resistance  may  then  be 
inserted  in  the  rotor  circuits.     Such  a  motor  is  very  expensive. 

403.  Cascade  Control  or  Concatenation. — When  operating  in 
cascade  two  similar  induction  motors  with  wound  rotors  are 
rigidly  connected  to  the  same  shaft.     The  stator  of  the  first 
motor  is  connected  to  the  line ;  the  stator  of  the  second  motor  is 
connected  to  the  rotor  winding  of  the  first  motor  and  receives 


INDUCTION  MOTOR  477 

power  from  it;  the  rotor  of  the  second  motor  is  closed  through 
starting  resistances  (Fig.  450). 

The  frequency  of  the  e.m.fs.  generated  in  the  rotor  of  an  in- 
duction motor  is  s/,  where  /  is  the  frequency  of  the  supply  and  s 
is  the  slip.  Thus  the  frequency  impressed  on  the  stator  of  the 
second  motor  is  the  frequency  of  slip  of  the  first  motor.  The 
speed  of  the  two  motors  is  always  the  same  and  thus  at  no  load 
(1  -  s)/  =  sf  and  s  =  0.5. 


FIG.  450. — Cascade  control  of  induction  motors. 

Therefore,  two  similar  motors  connected  in  cascade  tend  to 
approach  a  speed  of  half  synchronous  speed  at  no  load  and  fall 
below  this  speed  under  load.  Speeds  below  half  synchronous 
speed  are  obtained  by  inserting  resistance  in  the  rotor  windings 
of  the  second  motor.  For  speeds  above  half  synchronous  speed 
the  stator  of  the  second  motor  must  be  connected  to  the  line  and 
the  rotor  of  the  first  motor  closed  through  resistances. 

This  method  of  control  is  used  for  some  three-phase  traction 
systems  and  is  very  similar  to  the  series  parallel  control  of  direct- 
current  series  motors.  The  induction  motor  does  not,  however, 
tend  to  increase  its  speed  indefinitely.  If  it  operates  above  syn- 
chronous speed,  it  becomes  a  generator  and  feeds  back  power  to 
the  line  acting  as  a  brake. 

If  the  two  motors  have  different  numbers  of  poles,  the  no-load 
speed  when  connected  in  cascade  will  not  be  half  synchronous 
speed  but  the  speed  of  the  set  will  be  that  of  a  motor  having  as 
many  poles  as  the  two  motors  combined.  For  example,  if  on  a 
60-cycle  system  a  12-pole  motor  is  connected  in  cascade  with  a 
4-pole  motor  connected  to  run  in  the  same  direction  the  speed 

120  X  60 
will  be  that  of  a  motor  with  12  +  4  =  16  poles,  that  is, ^ — 

=  450  r.p.m.  When  the  two  motors  tend  to  run  in  the  same 
direction  they  are  in  direct  concatenation.  If  the  4-pole  motor 
is  connected  to  run  in  the  opposite  direction  from  the  12-pole 


478  ELECTRICAL  ENGINEERING 

motor,  the  resulting  speed  will  be  that  corresponding  to  12  —  4  = 
8  poles.  This  is  called  differential  concatenation. 

With  this  combination  of  motors,  four  speeds  may  be  obtained, 

120  X  60 

450  r.p.m.   with  direct  concatenation,   :  -  =  600  r.p.m. 

i& 

120  X  60 
using  only  the  12-pole  motor,  — io_4     =  900  r.p.m.   with  dif- 

120  X  60 
ferential   concatenation   and   finally   :    —r—  -  =  1,800   r.p-m., 

using  only  the  4-pole  motor.  Intermediate  speeds  can  be  ob- 
tained inserting  resistances  in  the  rotors. 

When  connected  in  cascade  the  impedance  of  the  second  motor 
is  added  to  the  rotor  of  the  first  and  its  exciting  current  is  supplied 
through  the  first  motor  resulting  in  very  low  power  factor.  With 
two  similar  motors  in  cascade  each  will  operate  at  about  half 
voltage  and  since  the  output  varies  as  the  square  of  the  voltage 
the  combined  output  of  the  set  will  be  about  half  that  of  each 
motor  separately.  The  losses  will  be  those  of  two  motors  and 
therefore  the  efficiency  will  be  very  low.  The  maximum  or 
breakdown  output  will  be  reduced  in  the  same  proportion. 

404.  Exciting  Current. — The  exciting  current  or  no-load  cur- 
rent of  an  induction  motor  is  larger  than  that  of  a  transformer 
and  is  of  the  order  of  25  per  cent,  of  full-load  current.  The 
power  component  supplies  the  iron  losses  of  the  stator  and  the 
friction  and  windage  losses  of  the  rotor.  The  quadrature  com- 
ponent or  magnetizing  current  is  large  because  the  magnetic  cir- 
cuit contains  an  air  gap  and  the  iron  on  both  sides  is  cut  away 
to  form  the  slots  for  the  rotor  and  stator  windings.  The  large 
magnetizing  current  lowers  the  power  factor  of  the  motor  par- 
ticularly at  light  loads  and  is  therefore  objectionable.  To  reduce 
it  the  air  gap  is  made  as  short  as  possible,  being  determined 
largely  by  the  necessity  for  mechanical  clearance.  The  slots  may 
be  partly  closed  to  increase  the  effective  gap  area;  the  rotor  slots 
are  often  completely  closed.  Closing  the  slots,  however,  adds  to 
the  reactance  of  the  motor. 

Low-speed  motors  have  larger  magnetizing  currents  than  high- 
speed motors  and  therefore  their  power  factors  tend  to  be  lower.  \ 

Unless  the  teeth  of  the  stator  and  rotor  are  properly  designed 
the  reluctance  of  the  magnetic  circuit  of  the  motor  will  vary  and 
the  flux  will  vary.  If  the  stator  and  rotor  have  the  same  number 
of  teeth,  then  in  position  (6),  Fig.  451,  the  reluctance  will  be  a 


INDUCTION  MOTOR 


479 


maximum  and  the  flux  a  minimum  while  in  (c)  the  flux  will  be 
maximum.  The  rotor  will  tend  to  lock  in  the  position  of  maxi- 
mum flux  (a)  and  the  torque  will  be  reduced.  Such  pulsations  of 
flux  may  be  prevented  by  designing  the  rotor  and  stator  with 
different  numbers  of  teeth  and  making  the  width  of  the  rotor 
teeth  a  multiple  of  the  stator  slot  pitch. 


(a)  Slot  Leakage  Flux 


(  C  )    Zig-Zag  Leakage  Flux 
FIG.  451. 

405.  Leakage  Reactances. — The  diameter  of  the  circle,  in  the 
diagram  for  the  induction  motor,  is  inversely  proportional  to  the 
sum  of  the  reactances,  and  it  is  therefore  necessary  for  the  de- 
signer to  determine  these  quantities  very  accurately  and  they 
should  ordinarily  be  kept  as  low  as  possible. 

The  leakage  fields  are:  (1)  that  about  the  end  connections,  (2) 
the  slot  leakage,  (3)  the  zig-zag  or  tooth-tip  leakage  and  (4)  the 
belt  leakage. 


480  ELECTRICAL  ENGINEERING 

The  end-connection  flux  may  be  reduced  by  using  fractional- 
pitch  windings  but  this  results  in  increased  magnetizing  current. 

The  slot  leakage  flux,  Fig.  451,  is  directly  proportional  to  the 
slot  depth  and  is  inversely  proportional  to  the  slot  width.  Deep 
narrow  slots,  therefore,  give  large  slot  leakage.  When  the  slots 
are  partly  closed,  to  decrease  the  magnetizing  current,  the  slot 
leakage  is  very  much  increased.  The  use  of  magnetic  wedges  in 
the  slots  gives  the  same  result. 

The  zig-zag  leakage  flux,  Fig.  451(6),  corresponds  to  the  tooth- 
tip  leakage  flux  in  alternators  and  is  usually  quite  large  since  the 
gap  is  short.  It  also  depends  on  the  relative  position  of  the  rotor 
and  stator  teeth.  In  position  (6),  Fig.  451,  the  leakage  flux  is 
maximum  and  in  (c)  it  is  minimum.  Such  a  large  pulsation  of 
the  flux  would  have  very  serious  results.  The  rotor  would  tend 
to  lock  in  the  position  of  maximum  flux  especially  when  being 
started  at  reduced  voltage  with  large  current.  The  rapid  pulsa- 
tion of  the  flux  might  also  produce  an  objectionable  noise. 

If  the  numbers  of  teeth  on  the  stator  and  rotor  have  no  common 
divisor,  only  one  tooth  will  be  in  the  position  of  maximum  flux  at 
any  time  and  the  tendency  to  lock  will  be  negligible.  If  less  than 
20  per  cent,  of  the  teeth  are  in  the  position  of  maximum  flux  at 
any  time  the  tendency  to  lock  will  not  be  serious.  This  condition 
can  easily  be  fulfilled  in  a  squirrel-cage  rotor  but  with  wound 
rotors  the  number  of  slots  on  the  rotor  and  stator  must  be  divis- 
ible by  the  number  of  poles  and  phases  and  conditions  are  not 
so  good.  However,  wound-rotor  machines  are  usually  started 
with  resistance  in  the  rotor  circuits  and  the  starting  current  and 
the  zig-zag  leakage  are  therefore  small. 

The  belt-leakage  flux  is  due  to  the  fact  that  the  stator  phases 
are  not  fixed  in  position  relative  to  the  rotor  phases  as  in  the  trans- 
former and  increased  leakage  results  when  they  are  not  directly 
opposite. 

When  the  sum  of  the  reactances  of  a  motor  is  small,  the  diame- 
ter of  the  circle  is  large  and  the  pull-out  torque  is  large  but  the 
starting  current  is  also  high. 

406.  Stator  and  Rotor  Resistances. — The  stator  resistance  n 
causes  a  loss  of  power  and  should  be  kept  as  small  as  possible. 
The  rotor  resistance  r2  likewise  causes  a  loss  of  power  but  it  may 
be  necessary  to  make  it  comparatively  large  in  order  to  get  suit- 
able characteristics. 

The  slip  is  proportional  to  the  rotor  copper  loss  and  therefore 


INDUCTION  MOTOR  481 

for  low  slip  the  rotor  resistance  must  be  low.  However,  by 
increasing  the  rotor  resistance  the  starting  torque  is  increased  and 
the  starting  current  is  reduced  and  where  large  starting  torque 
is  required  a  comparatively  high  rotor  resistance  is  necessary. 

With  wound  rotors  external  resistance  may  be  introduced  at 
start  and  then  cut  out,  but  with  the  squirrel-cage  rotor  the  resist- 
ance cannot  be  varied. 

If  the  starting  torque  of  a  squirrel-cage  rotor  is  found  to  be  too 
small,  it  may  be  increased  by  slotting  the  rotor  end  rings  and 
so  increasing  the  resistance. 

If  the  bars  of  a  squirrel-cage  rotor  are  made  very  deep,  good 
starting  torque  can  be  obtained  without  reducing  the  efficiency. 
At  start  the  frequency  of  the  rotor  current  is  high  and  the  slot 
leakage  flux  is  large,  the  lower  parts  of  the  bars  have  a  high 
reactance  and  carry  very  little  current.  Due  to  the  unequal 
current  distribution  the  effective  rotor  resistance  is  increased  and 
the  Pr  loss  is  increased  directly  increasing  the  starting  torque. 
Near  synchronous  speed  the  rotor  current  and  frequency  both 
become  low  and  the  current  is  uniformly  distributed  and  the 
effective  rotor  resistance  approaches  its  true  value.  The  effi- 
ciency is  therefore  not  reduced  but  due  to  the  high  slot  leakage 
the  reactance  is  increased  and  the  maximum  output  is  decreased. 

407.  Effect  of  Change  of  Voltage  and  Frequency. — If  the 
voltage  impressed  on  an  induction  motor  is  decreased  10  per 
cent,  while  the  frequency  is  maintained  constant,  the  flux  is 
decreased  10  per  cent,  and  the  exciting  current  is  decreased  more 
than  10  per  cent. ;  the  iron  loss  is  decreased  nearly  20  per  cent. ; 
the  diameter  of  the  circle  is  decreased  10  per  cent.  For  a  given 
output  the  current  is  increased  10  per  cent,  and  the  copper  loss  is 
increased  20  per  cent. ;  the  slip  is  increased  about  20  per  cent. ; 
the  speed  is  reduced.  The  efficiency  is  decreased  slightly.  The 
power  factor  at  light  load  is  increased  due  to  the  decreased  excit- 
ing current  but  above  full  load  it  is  decreased  due  to  the  increased 
current  and  decreased  circle  diameter.  The  maximum  output 
and  maximum  torque  are  decreased  20  per  cent. 

If  the  frequency  is  increased  10  per  cent,  while  the  voltage  is 
kept  constant,  the  flux  is  decreased  10  per  cent. ;  exciting  current 
is  decreased  more  than  10  per  cent. ;  the  iron  loss  remains  about  as 
before;  the  diameter  of  the  circle  is  decreased  10  per  cent.  For 
the  same  output  the  current  is  as  before ;  the  copper  loss  and  slip 
are  as  before  but  the  speed  is  higher  by  10  per  cent.  The  effi- 

31 


482 


ELECTRICAL  ENGINEERING 


ciency  is  as  before.  The  power  factor  is  improved  at  light  loads. 
Maximum  output  and  torque  are  decreased  less  than  10  per  cent. 

A  10  per  cent,  increase  in  voltage  would  be  pretty  well  offset 
by  a  10  per  cent,  decrease  in  frequency  but  the  increased  losses 
and  lower  speed  would  cause  the  motor  to  heat  up. 

408.  Single -phase  Induction  Motor.1— The  stator  of  a  single- 
phase  induction  motor  has  a  single  winding  with  any  number  of 
pairs  of  poles. 

The  rotor  is  either  of  the  squirrel-cage  type  or  is  wound  with 
the  same  number  of  poles  as  the  stator  but  with  any  number  of 
phases. 


Y 

FIG.  452.  FIG.  453.  FIG.  454.  FIG.  455.  FIG.  456. 

FIGS.  452  TO  456. — Single-phase  induction  motor. 

Fig.  452  shows  the  relative  directions  of  the  currents  in  the  two 
windings  at  standstill.  The  stator  carries  a  current  /i  which 
consists  of  two  components,  /'  the  load  component  and  IM  the 
magnetizing  current.  The  rotor  carries  a  current  72  opposite  in 
phase  to  /'  and  equal  to  it  in  m.m.f.  If  the  ratio  of  turns  is 
assumed  to  be  ft  i  :  nz  =  1  :  l;then/2  =  7'.  The  motor  at  stand- 
still is  a  transformer  with  a  short-circuited  secondary. 

The  flux  which  crosses  the  air  gap  and  links  with  both  stator 
and  rotor  is  produced  by  the  stator  exciting  current.  It  is 
always  directed  along  the  line  YOY.  There  is  no  component  of 
flux  in  the  horizontal  direction  XOX  and  therefore  no  torque  is 
exerted  tending  to  turn  the  rotor  in  either  direction.  Thus  the 
rotating  field  which  is  produced  in  the  polyphase  induction  motor 
does  not  exist  in  the  single-phase  motor  at  standstill.  The  single- 
phase  induction  motor,  therefore,  has  no  starting  torque.  If, 
however,  it  is  started  in  either  direction  it  will  develop  torque 
and  will  accelerate  and  come  up  approximately  to  synchronous 
speed  at  no  load. 

Fig.  453  represents  the  motor  with  the  rotor  open-circuited  and, 
therefore,  without  current  in  its  windings.  The  stator  carries 
only  the  magnetizing  current. 

Fig.  454  represents  conditions  at  synchronous  speed  at  the 


INDUCTION  MOTOR  483 

instant  when  the  stator  magnetizing  current  is  maximum.  The 
stator  flux  is  then  maximum  downwards. 

The  rotor  conductors  moving  at  synchronous  speed  cut  the 
stator  flux  and  an  e.m.f .  is  generated  in  them  proportional  to  the 
product  of  flux  and  speed.  Since  the  flux  is  alternating  the  e.m.f. 
generated  is  of  double  frequency  and  produces  a  current  of  double 
frequency  in  the  closed-rotor  winding.  The  current  produces  a 
flux  the  rate  of  change  of  which  through  the  rotor  windings  gener- 
ates in  them  an  e.m.f.  equal  and  opposite  to  the  e.m.f.  generated 
by  rotation.  This  flux  must,  therefore,  be  of  the  same  value  as 
the  stator  flux  and  it  is  in  phase  with  the  rotor  current. 

The  rotor  current  goes  through  two  complete  cycles  during  one 
revolution.  In  Fig.  454  it  is  maximum  and  is  opposed  to  the 
stator  current,  but  the  e.m.f.  impressed  on  the  stator  is  constant 
and  the  stator  flux  is  constant,  and,  therefore,  a  current  must  flow 
in  the  stator  to  balance  the  m.m.f .  of  the  rotor  current  I' M.  Since 
the  ratio  of  turns  has  been  taken  as  1  :  1  the  increase  in  stator 
current  is  I' M  and  the  total  stator  current  at  synchronous  speed  is 
IM  +  I'M.  In  the  position  shown  the  rotor  flux  is  not  produced 
because  the  rotor  m.m.f.  is  opposed  by  an  equal  and  opposite 
m.m.f.  on  the  stator. 

Fig.  455  represents  conditions  after  the  rotor  has  turned 
through  one-half  a  revolution  and  the  stator  current  has  passed 
through 'one-half  cycle.  The  rotor  current  is  in  the  same  direc- 
tion as  before  and  has  completed  one  cycle. 

Fig.  456  represents  conditions  midway  between  Fig.  454  and 
Fig.  455.  The  stator  current  is  zero  and  the  rotor  current  is 
maximum  and  exerts  a  m.m.f.  in  the  horizontal  direction.  There 
is  no  stator  m.m.f.  opposing  it  and  a  flux  is  produced  of  the  same 
value  as  the  stator  flux  in  Fig.  454  or  Fig.  455.  Since  the  reluc- 
tance of  the  path  for  the  horizontal  flux  is  the  same  as  that  for  the 
vertical  stator  flux,  the  rotor  magnetizing  current  I'M  must  be 
equal  to  the  stator  magnetizing  current  at  standstill  IM,  and, 
therefore,  at  synchronous  speed  the  stator  magnetizing  current 
is  21  M  and  is  double  its  value  at  standstill. 

Thus  at  synchronous  speed  there  is  a  resultant  m.m.f.  of  con- 
stant value  revolving  at  synchronous  speed  and  the  magnetic 
field  of  the  single-phase  motor  is  identical  with  that  of  the  poly- 
phase motor,  Fig.  457.  The  m.m.f.  to  produce  the  vertical  field  is 
supplied  by  the  true  stator  magnetizing  current,  while  the  m.m.f. 
to  produce  the  horizontal  field  is  provided  by  an  equal  stator 


484 


ELECTRICAL  ENGINEERING 


magnetizing  current,  in  phase  with  the  true  stator  magnetizing 
current,  which  induces  in  the  rotor  the  rotor  magnetizing  current. 

When  the  rotor  runs  at  synchronous  speed  its  conductors  do 
not  cut  this  revolving  flux  and  the  only  current  in  the  rotor  is  the 
double-frequency  magnetizing  current. 

When  the  rotor  runs  at  a  slip  s  below  synchronous  speed  the 
rotor  conductors  cut  the  flux  and  currents  are  produced  in  them 
and  torque  is  developed  just  as  in  the  case  of  the  polyphase  motor. 


FIG.  457. — Revolving  field  of  a 
single-phase  induction  motor  at 
synchronous  speed. 


FIG.  458. — Revolving  field  of  a 
single-phase  induction  motor  at 
slip  s. 


409.  Horizontal  Field  at  Slip  s. — When  the  rotor  runs  at  a 
speed  S  =  (I  —  s)  X  synchronous  speed,  the  e.m.f.  generated  in 
it  due  to  cutting  the  stator  flux  is  less  than  at  synchronous  speed 
in  the  ratio  1  —  s  :  1  and  the  horizontal  flux  and  the  rotor  mag- 
netizing current  are  less  in  the  same  ratio. 

The  stator  current  is  IM  +  (1  —  s)  IM  +  I'  and  the  rotor  cur- 
rent is  (1  —  s)  IM  +  72.  The  frequency  of  the  rotor  magnetizing 
current  is  (2  —  s)f  and  the  frequency  of  the  rotor  load  current  is 
sf,  where  /  is  the  frequency  of  the  e.m.f.  impressed  on  the  stator. 

The  revolving  field  at  slip  s  is  not  constant  in  value  but  has  the 
horizontal  axis  shorter  than  the  vertical  in  the  ratio  1  —  s  :  1, 
Fig.  458.  The  field  follows  an  elliptical  instead  of  a  circular  locus. 

The  torque  which  is  proportional  to  the  product  of  the  rotor 
load  current  and  the  horizontal  field  is  less  than  that  produced  in 
the  polyphase  motor  in  the  ratio  1  —  s  :  1. 

410.  Starting  Single-phase  Induction  Motors. — In  order  to 
obtain  the  torque  required  to  start  a  single-phase  induction  motor 
a  component  of  flux  in  quadrature  in  time  and  in  space  with  the 
stator  flux  must  be  produced  at  standstill.     It  has  been  shown 
that  when  once  the  motor  is  started  the  rotor  produces  the 
required  quadrature  flux  and  thus  the  torque  to  carry  the  load. 


INDUCTION  MOTOR 


485 


Two  principal  methods  are  employed  to  produce  the  quadrature 
flux  at  standstill:  (1)  phase  splitting  and  (2)  shading  coils. 

1.  If  the  two  stator  windings  of  a  two-phase  induction  motor 
are  connected  to  a  single-phase  supply,  phase  1  directly  and  phase 
2  through  a  suitable  resistance  or  condensive  reactance,  the  flux 
produced  by  phase  2  will  have  a  component  in  quadrature  in  time 
with  phase  1  and  will  thus  give  the  required  starting  torque, 
Fig.  459.  When  the  motor  has  come  up  to  half  speed,  the  starting 
winding  is  cut  out  and  the  motor  runs  as  a  single-phase  motor  on 
phase  1.  This  method  of  starting  is  called  phase  splitting.  The 
second  winding  need  not  have  as  many  turns  as  the  first  but  it 
should  be  placed  at  90  electrical  degrees  to  it. 

Fig.  459(6)  shows  a  three-phase  motor  connected  to  a  single- 
phase  circuit.  Two  phases  are  used  for  normal  operation  and 
the  third  phase  is  used  for  starting  only. 


Stator  Pole 


Starting 
Wlndi 


(a) 


Starting 
Winding 


FIG.  459. 


FIG.  460. 


2.  The  shading  coils,  Fig.  460,  are  short-circuited  coils  surround- 
ing part  of  each  pole  of  the  stator.  Currents  are  induced  in  them 
and  oppose  the  increase  and  decrease  of  the  flux  in  the  parts  of  the 
poles  which  they  inclose.  Thus  the  north  pole  in  section  A  will 
reach  its  maximum  value  before  it  is  maximum  in  section  B. 
When  the  north  pole  has  decreased  to  zero  in  B  it  will  be  in- 
creasing in  section  C  and  thus  there  is  a  rotation  of  the  mag- 
netic field  and  torque  is  produced.  When  the  motor  is  started 
the  short-circuited  coils  may  be  opened  and  they  will  then  be  idle 
and  will  not  cause  any  power  loss. 

411.  Comparison  of  Single-phase  and  Polyphase  Motors. — 
Take  the  case  of  a  two-phase  motor  operating  on  a  single-phase 
circuit  using  only  one  phase  of  the  stator  winding. 

The  slip  single-phase  is  less  than  two-phase  since  the  whole 
rotor  corresponds  to  one  phase  of  the  stator  and  thus  the  rotor 
current  and  rotor  copper  loss  are  decreased. 

The  efficiency  is  lower  because  the  output  decreases  more  than 


486 


ELECTRICAL  ENGINEERING 


i60 

§50 
| 

£40 


the  losses.     For  a  given  impressed  e.m.f.  and  frequency  the  iron 
and  friction  losses  remain  practically  constant. 

The  power  factor  is  lower  because  the  magnetizing  current  is 
approximately  doubled. 

A  given  motor  wound  single-phase  can  be  operated  at  higher 
densities  than  when  wound  polyphase  since  the  losses  are  less 
and  its  ventilation  is  the  same,  and  in  this  way  its  output  may  be 
made  from  65  to  75  per  cent,  of  its  output  polyphase. 

The  torque  at  any  speed  can  be  increased  by  introducing  re- 
sistance into  the  rotor  windings, 
but  this  changes  the  maximum 
torque  since  the  torque  is  propor- 
tional to  1  —  s. 

Fig.  461  shows  typical  speed- 
torque  curves  of  a  single-phase 
induction  motor  with  various 
external  resistances  connected  in 
the  rotor  windings. 

A  single-phase  induction  mo- 
tor is  usually  either  a  two-phase 
or  three-phase  motor  operated 
on  a  single-phase  circuit  using 
only  part  of  the  stator  winding. 
If  one  phase  of  a  two-phase 
motor  is  opened  at  light  load,  the  magnetizing  current  of  the 
other  phase  is  doubled  and  the  motor  runs  as  a  single-phase 
motor.  If  two  phases  of  a  three-phase  motor  are  opened  the 
motor  runs  as  a  single-phase  motor  with  the  magnetizing  current 
in  the  third  phase  trebled.  In  both  cases  the  flux  distribution 
and  flux  densities  remain  approximately  the  same  as  before. 

412.  Induction  Generator. — If  the  stator  of  an  induction 
motor  is  connected  to  the  supply  lines  and  its  rotor  is  driven 
above  synchronous  speed,  the  machine  will  develop  electrical 
power  and  supply  it  to  the  system. 

The  stator  flux  is  not  affected  by  the  increase  in  the  speed  of 
the  rotor,  but  revolves  in  the  same  direction  as  when  the  machine 
operates  as  a  motor.  The  slip  is,  however,-  reversed  and  the 
e.m.fs.  and  currents  induced  in  the  rotor  are  reversed.  Thus 
the  direction  of  torque  and  power  is  reversed  and  the  mechanical 
power  supplied  to  drive  the  rotor  is  transformed  into  electrical 
power  and  supplied  over  the  lines  to  the  load. 


0       20     40      60      80     100     120    140    100   180 
Torque  in  Percent  of  Full  Load 

FIG.  461. — Speed-torque  curves  of  a 
single-phase  induction  motor. 


INDUCTION  MOTOR 


487 


The  power  transferred  from  the  rotor  to  the  stator  depends  on 
the  slip  just  as  in  the  induction  motor.  Using  the  same  notation 
as  in  Art.  386,  the  power  transferred  to  the  stator  is 

P  =  nEzh  cos  62 


7-22  +  s  V 

Thus  to  increase  the  power  delivered  by  the  generator  its  speed 
must  be  increased.  If  therefore  an  induction  generator  is  con- 
nected to  a  prime  mover  of  variable  speed,  it  will  supply  power 
almost  in  proportion  to  the  increase  of  its  speed  above  synchro- 
nous speed. 

The  characteristics  of  an  induction  generator  may  be  deter- 
mined from  a  circle  diagram  (Fig.  462).  The  diagram  is  con- 
structed as  for  a  motor  but  since  the  slip  is  negative  the  complete 
circle  is  required.  Points  below  the  line  oa  represent  negative 
slip  and  therefore  power  output  as  a  generator. 


FIG.  462. — Circle  diagram  for  an  induction  generator. 

Assuming  the  generator  to  be  three-phase  star-connected,  the 
voltage  per  phase  is  E0  and  the  voltage  between  terminals  is 
E  =  V3#G. 

For  the  point  d: 

Input  =  \/3E  X  dm  watts  =  mechanical   power 

supplied. 
Rotor  copper  loss  =  \/3E  X  pm  watts. 


488 


ELECTRICAL  ENGINEERING 


Output  to  stator  =  \/3E  X  dp  watts 
Stator  copper  loss  =  \/3E  X  qp  watts. 
Stator  core  loss  =  \/3E  v  ™  watts 
Output  =  A/2tf 

Efficiency 


qp  watts. 
X  rq  watts. 
X  dr  watts. 

j—  100  per  cent. 
dm 


The  magnetizing  current  oa  at  no  load  and  or  under  load  must 
be  supplied  by  a  synchronous  machine  operating  in  parallel  with 
the  induction  generator  (Fig.  463).  The  voltage  and  the  fre- 
quency of  the  system  are  also 
fixed  by  this  synchronous  gen- 
erator and  are  not  affected  by 
the  induction  generator. 

The  induction  generator  has 
two  very  serious  disadvantages ; 
it  is  not  self-exciting,  and  it 
cannot  supply  reactive  currents 


Rotor 
Driven 
'abore 

Synchrom 
Speed 


Induction.  Generator 
FIG.  463. — Induction  generator. 


to  an  inductive  load.  It  must, 
therefore,  be  operated  in  parallel 
with  an  alternator  of  sufficient 
capacity  to  supply  both  the 

magnetizing  current  for  the  induction  generator  and  the  reactive 
currents  required  by  the  load.  The  exciting  kilovolt-amperes  at 
no  load  may  be  as  low  as  12  per  cent,  of  the  full-load  rating  but 
under  load  it  will  never  be  less  than  25  per  cent.  If,  however, 
the  output  of  the  induction  generator  is  small  in  comparison  with 
the  capacity  of  the  synchronous  machines  on  the  system  this  will 
not  have  serious  effects. 

The  induction  generator  has  some  good  points:  (a)  it  does  not 
require  to  be  synchronized ;  (6)  there  is  no  danger  of  trouble  due 
to  hunting,  since  it  does  not  operate  at  constant  speed ;  (c)  in  case 
of  short-circuits  it  loses  its  excitation  and  dogs  not  tend  to  supply 
power  to  the  fault. 

In  construction  an  induction  generator  is  similar  to  an  induc- 
tion motor  but  the  exciting  current  must  be  kept  as  low  as  pos- 
sible by  making  the  air  gap  short.  Since  no  starting  torque  is 
required,  the  rotor  may  be  made  of  very  low  resistance  and  is 
usually  of  the  squirrel-cage  type. 

413.  Asynchronous  Phase  Modifier. —The  asynchronous  phase 
modifier  or  phase  advancer  is  a  machine  for  use  with  induction 
motors  to  improve  their  power  factor ;  it  fills  the  place  of  the  ex- 


INDUCTION  MOTOR 


489 


citer  for  a  synchronous  motor.  A  number  of  different  types  have 
been  designed  but  only  the  simplest  one  will  be  discussed  here. 
It  consists  of  a  direct-current  drum  armature  with  a  commutator 
and  three  sets  of  brushes  per  pair  of  poles  displaced  at  120  degrees. 
The  stator  is  merely  a  frame  with  laminations  but  without  slots 
or  windings  except  in  the  larger  sizes  where  a  compensating  wind- 
ing may  be  required  to  take  care  of  commutation.  For  small 
sizes  the  stator  may  be  omitted  if  the  armature  windings  are 
placed  in  totally  closed  slots.  The  iron  above  the  slots  serves 
to  complete  the  magnetic  circuit. 


Transformer 


Induction 
Motor 
Horse-power 


Starting        Switches 
Resistances 

FIG.  464. — Asynchronous-phase  modifier. 

Referring  to  Fig.  464,  A  is  the  main  motor,  which  must  have  a 
wound  rotor  and  starting  resistances  R.  The  switch  S  is  used 
to  disconnect  the  phase  advancer  and  close  the  rotor  circuits 
while  starting  up.  When  full  speed  is  reached,  the  switch  S  is 
opened,  connecting  the  terminals  of  the  rotor  to  the  three  brushes 
on  the  phase  advancer  B.  If  the  armature  is  at  rest  it  acts  as  a 
three-phase  reactance  and  produces  a  revolving  field  of  the  fre- 
quency of  slip.  When  the  armature  is  driven  in  the  direction 
of  the  revolving  field  and  at  the  same  speed  the  reactance  be- 
comes zero.  If  the  speed  is  then  increased,  the  reactance  is 
reversed  and  acts  as  a  capacity  reactance  causing  the  rotor 
current  7  2  to  lead  the  rotor  induced  e.m.f.  E2o)  and  so  improves 
the  stator  power  factor. 

The  change  in  phase  is  illustrated  in  Fig.  465.  (a)  is  the  dia- 
gram without  the  phase  advancer  and  (b)  represents  the  case 
where  the  phase  advancer  is  driven  at  such  a  speed  that  the 
stator  current  1 1  is,  in  phase  with  the  impressed  e.m.f.  E\.  The 


490 


ELECTRICAL  ENGINEERING 


rotor  current  72  has  then  two  components,  I't  the  component  ex- 
erting a  m.m.f.  equal  and  opposite  to  that  of  the  stator  current 
/i  and  I'M,  the  rotor  magnetizing  current  filling  the  place  of 
the  ordinary  stator  magnetizing  current.  The  rotor  reactance 
drop  is  shown  to  be  reversed  and  is  approximately  sIi(xB  —  x2) 
where  XB  is  the  condensive  reactance  of  the  phase  advancer  wind- 
ing at  the  given  speed.  By  driving  the  phase  advancer  at  a  still 
higher  speed  the  primary  power  factor  may  be  made  leading. 


FIG.  465. 


The  capacity  of  the  phase  advancer  is  very  small,  being  equal 
to  the  required  change  in  the  stator  reactive  volt  amperes  multi- 
plied by  the  slip.  This  may  be  understood  from  the  following: 
A  stator  e.m.f.  E\Q  corresponds  to  a  rotor  e.m.f.  EZg  =  sE%  = 


—  Eig. 


A  stator  current  I'  corresponds  to  a  rotor  current  / 


—  /'.     And,  therefore,  the  stator  volt-amperes  Eigl'  correspond 

to  the  rotor  volt-amperes  E^glz  =  sEig!f. 

The  phase  advancer  is  driven  by  a  small  high-speed,  squirrel- 
cage  induction  motor  C  connected  to  the  supply  lines  directly  or 
through  a  transformer.  It  supplies  the  friction  windage  and  iron 
losses  of  the  phase  advancer  but  the  copper  loss  is  supplied  from 
the  rotor  of  the  main  motor. 


INDUCTION  MOTOR  491 

The  capacity  of  the  main  induction  motor  is  very  largely  in- 
creased by  the  elimination  of  the  reactive  currents  from  the 
stator.  The  supply  lines  are  also  relieved  of  a  large  component 
of  current. 

To  raise  the  power  factor  of  a  1,000-kva.  induction  motor  from 
86.6  per  cent,  to  100  per  cent.,  the  reactive  kilo  volt-amperes  re- 
quired in  the  stator  =  1,000  Vl  -0.866^=  1,000  X  0.50  =  500 
kva.  Assuming  a  slip  of  3  per  cent,  the  capacity  of  the  phase 
advancer  is  only  0.03  X  500  =  15  kva.  A  K-hp-  driving  motor 
would  supply  the  losses.  The  power  developed  by  the  induction 
motor  is  increased  from  1,000  X  0.866  =  866.0  kw.  to  1,000  kw., 
that  is,  by  about  15  per  cent. 

To  reduce  the  required  phase  advancer  capacity  the  slip  of  the 
motor  should  be  made  as  small  as  possible  by  decreasing  its  rotor 
resistance. 

If  the  equipment  of  a  shop  consists  of  one  large  motor  and  a 
number  of  smaller  ones,  a  phase  advancer  may  be  applied  to  the 
large  motor  to  make  its  power  factor  leading  and  so  supply  the 
reactive  currents  for  all  the  motors. 

414.  Phase  Converter. — To  change  from  one  polyphase  system 
to  another,  as  from  two-phase  to  three-phase  or  three-phase  to 
six-phase,  stationary  transformers  should  be  used;  but  when  a 
change  from  single-phase  to  two-phase  or  three-phase  is  required, 
transformers  are  no  longer  satisfactory  since,  due  to  the  pulsating 
nature  of  single-phase  power,  energy  storage  must  take  place 
and  this  can  best  be  accomplished  by  means  of  the  momentum  of 
a  revolving  machine. 

To  change  from  single-phase  to  two-phase  an  ordinary  two- 
phase  induction  motor  with  a  squirrel-cage  rotor  may  be  used 
as  a  phase  converter.  The  connections  are  shown  in  Fig.  466 (a). 
One  phase  A  of  the  stator  winding  is  connected  to  the  single- 
phase  supply  and  the  converter  runs  as  a  single-phase  induction 
motor  and  near  synchronous  speed  a  revolving  field  is  produced 
of  approximately  constant  value  (Art.  409).  This  field  cuts  the 
second  stator  phase  B  and  generates  in  it  a  voltage  in  quadrature 
with  the  supply  voltage  and  with  approximately  the  same  effect- 
ive value  at  no  load,  Fig.  466(6). 

One  phase  of  the  two-phase  system  receives  its  power  directly 
from  the  single-phase  supply  and  the  second  phase  is  supplied 
from  the  phase  B  of  the  converter.  Under  load  the  voltage  of 
phase  1  remains  constant  but  that  of  2  decreases  due  to  increased 


492 


ELECTRICAL  ENGINEERING 


slip  and  to  the  impedance  drop  in  the  converter  and  it  falls  back 
in  phase  to  Ef2)  Fig.  466  (c).  This  unbalance  of  the  two-phase 
system  may  be  corrected  for  a  given  load  by  changing  the  connec- 
tions as  indicated  by  the  broken  lines  in  Fig.  466 (a).  The  voltage 
across  phase  A  under  load  is  increased  by  changing  the  trans- 
former tap,  thus  raising  the  voltage  generated  in  B  and  in  addi- 
tion a  component  oa  of  the  supply  voltage  is  connected  in  series 
with  B.  The  voltage  of  phase  2  is  thus  changed  from  Efz  back 
to  E2. 


•        Single-Phas 

[000000000000 


Two-Pbase 


(a) 


Three-phase  Motor 


FIG.  466. 
FIGS.  466  AND  467. — Phase  converters. 


FIG.  467. 


When  changing  from  single-phase  to  three-phase  a  similar 
phase  converter  may  be  used  but  phase  B  must  be  wound  for 
only  87  per  cent,  of  the  voltage  of  phase  A.  Fig.  467 (a)  shows 
the  connections  required  for  operating  a  three-phase  induction 
motor  from  a  single-phase  circuit  and  the  equilateral  triangle 
abc  represents  the  balanced  three-phase  voltages  at  light  load. 
Under  load  the  voltage  EB  decreases  and  falls  back  in  phase  and 
the  point  c  moves  to  c'.  To  balance  the  system  the  connection 
for  phase  B  must  be  changed  from  d  to  df  and  the  voltage  im- 
pressed on  A  must  be  increased  by  changing  to  tap  bf.  This, 
however,  gives  a  balance  only  at  one  particular  load. 

In  both  cases  the  momentum  of  the  rotor  stores  and  returns 
energy  and  makes  possible  the  change  from  pulsating  single- 
phase  power  to  constant  polyphase  power. 

By  using  a  phase  converter  three-phase  induction  motors  can 
be  operated  from  a  single-phase  supply  circuit  for  the  propulsion 
of  electric  locomotives.  A  single  converter  may  be  used  to 
supply  a  number  of  induction  motors  and  it  may  be  designed  as  a 
high-speed  machine  and  will  therefore  be  light  and  cheap;  and 


INDUCTION  MOTOR  493 

since  it  exerts  very  little  torque  the  shaft  may  be  made  small. 
To  reduce  the  losses  and  slip  the  rotor  should  be  made  of  very 
low  resistance.  The  weight  of  the  phase  converter  may  be  only 
about  25  or  30  per  cent,  of  the  combined  weight  of  the  motors 
which  it  supplies. 

The  function  of  the  phase  converter  is  reversible  and  when  the 
motors  are  running  above  synchronous  speed  they  become  gen- 
erators supplying  single-phase  power  back  to  the  system  and 
give  a  large  and  uniform  braking  effect. 

415.  Induction  Frequency  Converter. — The  induction  fre- 
quency converter  may  be  used  instead  of  the  synchronous-motor 
generator  set  to  convert  power  from  one  frequency  to  another 
or  to  link  up  two  systems  of  different  frequencies.  It  consists 
of  an  induction  motor  with  a  wound  rotor  driven  by  a  synchro- 
nous motor  connected  to  the  25-cycle  supply  lines.  The  stator 
of  the  induction  motor  is  also  connected  to  the  supply  and 
produces  a  revolving  field.  At  standstill  the  frequency  of  the 
e.m.fs.  generated  in  the  rotor  windings  is  25  cycles.  When 
the  rotor  is  driven  backward  at  synchronous  speed  the  fre- 
quency is  50  cycles  and  when  driven  at  140  per  cent,  of  synchro- 
nous speed  it  is  60  cycles.  If  a  receiver  circuit  is  connected  to 
the  rotor  slip  rings  60-cycle  power  can  be  supplied  to  it. 
Twenty-five-sixtieths  or  five-twelfths  of  the  power  output  is 
supplied  to  the  rotor  by  transformer  action  from  the  stator  and 
the  remaining  seven-twelfths  is  supplied  by  the  synchronous 
motor  as  mechanical  power. 

The  principal  disadvantage  of  the  induction  frequency  con- 
verter is  its  poor  voltage  regulation.  Due  to  the  presence  of  the 
air  gap  in  the  magnetic  circuit  the  reactances  are  large  and  the 
e.m.fs.  consumed  by  the  reactances  are  large. 

The  exciting  current  required  by  the  induction  motor  may  be 
provided  by  over-exciting  the  fields  of  the  synchronous  motor 
and  making  it  draw  a  leading  current.  In  this  way  the  power 
factor  of  the  set  may  be  made  unity,  but  the  increased  current 
in  the  synchronous-motor  windings  increases  the  copper  losses 
and  the  heating. 

Since  the  rotor  of  an  induction  motor  may  be  wound  for  any 
number  of  phases  irrespective  of  the  number  of  stator  phases  the 
induction  frequency  converter  may  be  used  to  change  the  number 
of  phases  as  well  as  the  frequency.  The  same  result  may  of 
course  be  obtained  with  the  synchronous-motor  generator  set. 


CHAPTER  XIII 
ALTERNATING-CURRENT  COMMUTATOR  MOTORS 

416.  Motor   Characteristics. — Direct-current    motors    are    of 
three  types,  shunt,  compound  and  series.     The  methods  of  vary- 
ing the  speed  of  such  motors  were  discussed  in  Arts.  161  and  172. 

Of  the  alternating-current  motors,  the  synchronous  motor  runs 
at  a  constant  speed  at  all  loads  and  this  speed  cannot  be  varied. 
The  induction  motor  with  a  low-resistance  rotor  also  runs  ap- 
proximately at  constant  speed  and  this  can  be  decreased  only  by 
using  a  wound  rotor  with  slip  rings  and  inserting  resistance  in 
the  rotor  windings  to  increase  the  slip.  This  is  analogous  to  the 
introduction  of  resistance  in  series  with  the  armature  of  a  direct- 
current  motor  and  is  inefficient.  There  is  no  way  of  increasing 
the  speed  above  synchronous  speed  corresponding  to  the  field 
control  in  a  shunt  motor. 

The  single-phase  synchronous  motor  and  induction  motor  can- 
not start  under  load  but  the  polyphase  motors  exert  large  starting 
torque.  The  current  required,  however,  is  greater  than  for  a 
similar  direct-current  motor  since  it  is  not  in  phase  with  the  im- 
pressed voltage  and  the  torque  per  ampere  is  therefore  less. 

The  induction  motor  with  a  comparatively  high-resistance 
rotor  has  characteristics  very  similar  to  those  of  the  direct-current 
compound  motor.  The  speed  falls  off  under  load  and  the  start- 
ing torque  is  good. 

To  design  an  alternating-current  motor  with  characteristics 
similar  to  the  direct-current  series  motor  a  commutator  and 
brushes  must  be  added.  A  large  number  of  motors  of  this  kind 
have  been  developed  and  the  principles  of  their  operation  are 
discussed  below. 

By  adding  a  commutator  and  brushes  to  the  rotor  of  a  single- 
phase  induction  motor  the  power  factor  may  be  compensated  and 
the  speed  may  be  made  adjustable  from  values  above  synchro- 
nous speed  down  to  half  synchronous  speed. 

417.  Alternating-current  Series  Motor. — The  alternating-cur- 
rent series  motor  is  very  similar  to  the  direct-current  series  motor 
and  can  be  operated  on  direct-current  with  increased  efficiency 
and  output. 

494 


ALTERNATING-CURRENT  COMMUTATOR  MOTORS  495 

If  a  direct-current  series  motor  is  connected  to  an  alternating- 
current  supply  circuit  it  will  rotate  since  the  currents  in  the  field 
and  armature  reverse  together  and  therefore  the  torque  is  always 
in  one  direction,  but  it  will  be  very  inefficient  and  will  spark  very 
badly. 

With  alternating  current  flowing  in  the  field  winding  an  alter- 
nating magnetic  flux  is  set  up  through  the  magnetic  circuit  and 
causes  very  large  losses  due  to  hysteresis  and  eddy  currents.  To 
reduce  these  to  a  minimum  the  whole  magnetic  circuit  of  an 
alternating-current  series  motor  must  be  laminated.  The  field 
circuit  must  be  very  heavily  insulated  to  prevent  short-circuits 
between  turns  which  would  burn  out  the  motor  on  account  of 
the  large  induced  currents. 

The  relation  between  the  e.m.fs.  and  current  in  the  direct- 
current  series  motor  is  given  by  the  equation 

E  =  8  +  I(ra  +  TV),  (444) 

where   E  =  impressed  e.m.f., 

8   =  counter  e.m.f.  generated  by  rotation, 

/   =  current, 

ra  =  resistance  of  the  armature, 

r/  =  resistance  of  the  field. 

In  the  alternating-current  series  motor  the  alternating  flux  sets 
up  large  e.m.fs.  of  inductance  in  both  the  field  and  armature 
windings,  which  consume  components  of  the  impressed  e.m.f.  in 
quadrature  ahead  of  the  current.  If  L/  is  the  inductance  of  the 
field  and  La  the  inductance  of  the  armature,  their  reactances  are 
xf  =  27r/L/and  xa  =  2irfLa)  respectively,  where  /  is  the  frequency 
of  the  impressed  e.m.f. 


ry  a  Current    I 

FIG.  468. — Vector  diagrams  of  a  single-phase  series  motor. 

Fig.  468  shows  the  vector  diagram  for  the  motor. 

ox  =  I      =  current  in  field  and  armature. 

oa  =  Irf  =  e.m.f.  consumed  by  the  resistance  of  the  field. 


496  ELECTRICAL  ENGINEERING 

ab  =  Ixf  =  e.m.f.  consumed  by  the  reactance  of  the  field. 

be  =  Ira  =  e.m.f.  consumed  by  the  resistance  of  the  armature. 

cd  =  Ixa  =  e.m.f.  consumed  by  the  reactance  of  the  armature. 

dk  =  8  =  e.m.f.  generated  in  the  armature  due  to  rotation, 
in  phase  with  the  field  flux  and,  therefore,  in  phase  with 
the  current,  neglecting  the  hysteretic  lag. 

ok  =  E  =  impressed  e.m.f. 

cos  kox  =  cos  cj>  =  load  power  factor. 

cos  dox  =  cos  <$>s  =  power  factor  at  start. 

Taking  the  current  as  the  real  axis  the  relation  between  the 
current  and  the  impressed  e.m.f.  can  be  expressed  in  rectangular 
coordinates  as 

g  =  g  +  I(ra  +  rf)  +  jl(xa  +  xf)  (445) 

and  taking  absolute  values 


E  =       {&  +  I(ra  +  TV))2  +  \I(Xa  +  z,))2          (446) 
At  standstill 

E  =  lV(ra  +  rfy  +  (xa  +  xJY'  (447) 

and  the  current  is 

/  =  E  (448) 

+  rfy  +  (xa  +  xfY 


Full  voltage  can  usually  be  impressed  on  the  motor  at  stand- 
still without  causing  any  injury  since  the  current  is  limited  by 
the  large  impedance. 

The  power  factor  under  running  conditions  is 


cos  0  = 

but  8  =  kn$,  where  <i>  is  the  maximum  value  of  the  flux  per  pole, 
n  is  the  motor  speed  in  revolutions  per  second  and  k  is  a  constant 
depending  on  the  number  of  turns  in  the  armature  winding  and 
on  the  shape  of  the  flux  wave.  The  flux  $  is  almost  propor- 
tional to  the  current  I  and  the  generated  e.m.f.  may  be  expressed 
as 

8  =  k'nl. 

Substituting  this  value  for  8  in  equation  and  eliminating  I 

k'n  +  ra  +  r/  ,.  _n, 

cos  $  =      ,  J  =;  (450) 

V(k'n  +  ra  +  rfy  +  (xa  +  */)» 

the  power  factor,  therefore,  increases  with  increasing  speed  and 
approaches  unity.  At  low  speed  and  at  standstill  it  is  low  on 


ALTERNATING-CURRENT  COMMUTATOR  MOTORS  497 

account  of  the  reactances  in  the  field  and  armature  and  for  satis- 
factory operation  it  is  necessary  to  make  these  reactances  as  low 
as  possible. 

418.  Design  for  Minimum  Reactance. — The  inductance  of  any 
coil  is  proportional  to  the  square  of  the  number  of  turns  and  is 
inversely  proportional  to  the  reluctance  of  the  magnetic  circuit 
through  it.  To  reduce  the  inductance  L/  of  the  field  winding  it 
is  designed  with  a  small  number  of  turns  but  this  reduces  the 
field  m.m.f .  and  in  order  to  obtain  the  required  flux  the  reluctance 
of  the  magnetic  circuit  must  be  made  very  low.  For  this  purpose 
large  sections  of  high  permeability  are  used,  the  slots  are  partially 
closed  and  the  air  gap  is  made  as  short  as  possible. 

The  reactance  of  the  winding  is  proportional  to  the  product  of 
the  inductance  and  the  frequency  and  therefore  the  frequency 
should  be  low.  Motors  are  usually  designed  for  25  cycles  since 
that  is  the  lowest  standard  frequency,  but  they  will  operate  on 
15  cycles  or  on  direct  current  with  a  much  improved  efficiency 
and  power  factor  and  a  larger  output.  The  frequency  of  the 
supply  does  not  affect  the  speed  of  the  motor  directly,  but  it 
does  indirectly  since  the  reactance  drop  decreases  with  the  fre- 
quency and,  therefore,  the  speed  for  a  given  current  increases. 


FIG.  469. — Four-pole,   single-phase,   series  motor  with  compensating   winding. 

419.  Compensating  Windings. — The  armature  inductance  and 
reactance  cannot  be  decreased  by  reducing  the  number  of  turns 
on  the  armature  since,  for  a  given  impressed  voltage,  that  would 
increase  the  speed  of  the  motor,  and,  further,  since  the  field  is 
made  comparatively  weak  the  armature  must  be  made  corre- 

32 


498 


ELECTRICAL  ENGINEERING 


spondingly  strong  in  ampere-turns  in  order  to  produce  the  re- 
quired torque. 

The  armature  m.m.f.  as  in  direct-current  machines  is  cross- 
magnetizing  and  distorts  the  main  field  and  so  weakens  it  and 
interferes  with  commutation.  The  flux  produced  by  it  is  alter- 
nating and  induces  in  the  armature  a  back  e.m.f.  of  armature 
inductance.  This  flux  of  armature  reaction  or  armature  induct- 
ance may  be  reduced  as  in  the  direct  current  generator  by  the 
use  of  a  compensating  winding.  The  compensating  winding  is 
placed  in  slots  in  the  pole  faces  as  shown  in  Fig.  469.  It  is  dis- 
tributed over  the  whole  periphery  of  the  armature  and  exerts  a 
m.m.f.  opposing  the  armature  m.m.f.  and  so  limiting  the  cross 
flux  to  a  very  small  value  and  reducing  the  armature  inductance 
and  reactance  in  the  same  proportion. 

The  m.m.f.  of  the  compensating  winding  can  be  produced  in 
two  ways  illustrated  in  Fig.  470  and  Fig.  471.  The  first  is  called 
inductive  compensation  and  the  second  conductive  compensation. 


Field 
Winding, 


Variable  Ratio  Transformer 


FIG.  470. — Induc- 
tively compensated 
series  motor. 


FIG.  471.— Con- 
ductively  compen- 
sated series  motor. 


FIG.  472.— Series 
motor  with  variable 
voltage  supply. 


1.  In  the  inductively  compensated  series  motor  the  compensat- 
ing m.m.f.  is  produced  by  short-circuiting  the  compensating  coil. 
It  then  acts  as  the  closed  secondary  of  a  transformer  of  which  the 
armature  is  the  primary.     The  m.m.f.  of  the  compensating  wind- 
ing is  almost  equal  to  the  m.m.f.  of  the  armature  but  can  never  be 
greater  than  it  and,  therefore,  over-compensation  is  not  possible. 
The  combined  reactance  of  the  armature  and  compensating  wind- 
ing corresponds  to  the  reactance  of  a  transformer  on  short-circuit. 
The  mutual  flux  is  almost  destroyed  but  the  leakage  fluxes  remain. 

2.  In  the  conductively  compensated  motor  the  compensating 
coil  is  connected  in  series  with  the  field  and  armature  and  the 
amount  of  compensation  can  be  varied.     When  the  m.m.f s.  of 
the  two  windings  are  equal  there  is  no  mutual  flux  and  the  com- 
bined reactance  is  a  minimum.     When  the  m.m.f.  of  the  compen- 
sating winding  is  stronger  than  that  of  the  armature  the  armature 
reaction  flux  is  reversed  but  the  reactance  of  the  compensating 


ALTERNATING-CURRENT  COMMUTATOR  MOTORS  499 

winding  is  increased  and  so  part  of  the  advantage  is  lost,  but  the 
flux  due  to  over-compensation  assists  commutation  of  the  load 
current  in  the  same  way  that  interpoles  do  and  is  thus  a  great 
advantage. 

A  conductively  compensated  motor  can  be  operated  on  direct 
current  but  an  inductively  compensated  motor  cannot  since  the 
compensating  winding  would  not  be  effective  and  sparking  would 
occur. 

420.  Commutation. — Satisfactory  commutation  is  very  much 
more  difficult  to  obtain  in  the  alternating-current  series  motor 
than  in  the  direct-current  motor  because,  as  may  be  seen  in  Fig. 
473,  the  short-circuited  coil  is  in  the  position  of  the  short-circuited 
secondary  of  a  transformer  with  the  main  field  as  primary  and 
tends  to  have  as  many  ampere-turns  induced  in  it  as  there  are  on 


FIG.  473.  FIG.  474. 

a  pair  of  field  poles.  This  large  short-circuit  current  interferes 
with  commutation  and  must  be  reduced  as  far  as  possible.  For 
this  purpose  high-resistance  leads,  called  preventive  leads,  are 
connected  between  the  coils  and  the  commutator  bars,  as  shown 
in  Fig.  474,  and  narrow  carbon  brushes  of  high  contact  resistance 
are  used.  The  short-circuit  current  must  pass  through  two  resist- 
ance leads  in  series  and  is  thus  greatly  reduced  while  the  load 
current  is  carried  by  two  or  more  in  multiple.  The  resistance  of 
one  of  the  leads  must  be  very  much  higher  than  that  of  an  arma- 
ture coil  in  order  to  reduce  the  current  sufficiently. 

There  are  losses  in  the  leads  due  to  the  resultant  of  the  two 
currents  flowing  in  them,  but  by  increasing  the  resistance  up  to  a 
certain  point  the  short-circuit  current  is  reduced  and  the  com- 
bined loss  is  reduced. 

The  resistance  leads  are  not  made  of  large  enough  capacity  to 
carry  the  current  continuously  but  under  running  conditions  any 
one  lead  is  in  circuit  for  only  a  very  short  time.  If  the  motor 
is  stalled  with  power  on  the  leads  are  likely  to  be  destroyed. 

The  torque  of  the  motor  is  very  much  improved  by  the  use  of 


500 


ELECTRICAL  ENGINEERING 


resistance  leads  since  without  them  the  large  short-circuit  current 
would  weaken  the  main  field  and  decrease  the  torque. 

Fig.  475  shows  the  characteristic  curves  of  a  150  hp.,  single- 
phase  series  motor.  The  torque  and  speed  curves  are  very 
much  the  same  shape  as  those  of  the  direct-current  series  motor. 

The  power  factor  approaches  unity  at  light  load  when  the 
speed  is  high  as  explained  above,  but  at  full  load  it  is  still  very 
good,  reaching  90  per  cent,  in  some  cases.  At  start  and  at  low 
speeds  it  is  low  because  the  reactance  of  the  motor  is  constant. 


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Amperes 

Fio.  475. — Characteristic    curves    of    a    25-cycle,    250-volt,    150-horse-power, 
single-phase  series  motor. 

Efficiencies  up  to  85  per  cent,  can  be  obtained  but  the  motors 
must  be  designed  more  liberally  than  the  corresponding  direct- 
current  motors  and  are  therefore  heavier  and  more  expensive. 

On  account  of  unsatisfactory  commutation  alternating-cur- 
rent series  motors  are  only  built  for  voltages  of  250  volts  and 
under. 

421.  Speed  Control. — The  speed  of  the  series  motor  can  be 
controlled  by  supplying  the  motor  through  a  transformer  with 
a  number  of  secondary  taps  (Fig.  472).  By  changing  the  taps 
the  speed  of  the  motor  for  any  load  can  be  adjusted  through  a 
wide  range.  The  alternating-current  series  motor  has  in  this 
respect  an  advantage  over  the  direct-current  series  motor,  since 
the  change  of  impressed  voltage  is  not  accompanied  by  rheostatic 


ALTERNATING-CURRENT  COMMUTATOR  MOTORS  501 


1       " 

1     T 


422.  Polyphase  Commutator  Motor. — A  three-phase  commu- 
tator motor  with  a  series  characteristic  is  shown  in  Fig.  476.     The 
armature  is  similar  to  that  in  the  single-phase  motor  but  the  com- 
mutator has  three  sets  of  brushes  per  pair  of  poles,  spaced  at  120 
electrical   degrees.     The  three-phase  stator  winding  is  usually 
connected    to    the   line  through  trans- 
formers  and    the    other    terminals    are 

connected  to  the  three  brushes.  The 
machine  is  a  combination  of  three  single- 
phase  series  motors  in  one  and  has  similar 
characteristics.  If  three  compensating 
windings  are  added  to  the  stator,  at 
right  angles  to  the  three  exciting  wind- 
ings and  connected  in  series  with  them, 
the  power  factor  will  be  improved  but 
the  cost  will  be  increased.  The  speed 
for  any  load  can  be  adjusted  by  moving 
the  brushes  relative  to  the  exciting 
windings  but  the  120-degree  displace- 
ment must  be  maintained. 

423.  Repulsion  Motor. — In  construc- 
tion the  repulsion  motor  resembles  the 

single-phase  series  motor  with  conductive  compensation.  The 
armature  is,  however,  not  connected  in  series  with  the  field  but 
is  short-circuited  and  receives  its  current  by  induction. 

The  principle  of  its  operation  can  be  understood  by  reference 
to  Figs.  477  to  479.  In  Fig.  477  the  armature  is  shown  short- 
circuited  with  the  brushes  in  line  with  the  field  poles.  Current  is 


FIG.  476.— Three-phase 
commutator  motor. 


(a) 

FIG.  478.  FIG.  479. 

FIGS.  477  TO  479. — Repulsion  motor. 

induced  in  it  as  in  the  secondary  of  a  transformer  and  is  very 
large  but  the  torque  exerted  in  each  direction  is  the  same  and 
thus  the  resultant  torque  is  zero.  In  Fig.  478,  with  the  brushes 
turned  through  90  degrees  there  is  no  current  induced  in  the 
armature  and  therefore  no  torque.  In  order  that  the  motor  may 


502  ELECTRICAL  ENGINEERING 

exert  torque  the  brushes  must  be  placed  in  some  intermediate 
position  Fig.  479  (a).  The  same  result  is  accomplished  by  placing 
a  second  winding  at  right  angles  to  the  main  field  winding.  This 
is  shown  as  the  compensating  coil  C  in  Fig.  479(6)  and  is  carried 
in  slots  in  the  pole  faces  as  in  the  series  motor.  The  brushes  are 
placed  in  line  with  this  coil  and  the  armature  receives  its  current 
by  induction  from  it.  Torque  is  produced  which  is  proportional 
to  the  product  of  the  armature  current  induced  by  the  compensat- 
ing coil  and  the  flux  produced  by  the  main  field,  but  it  is  necessary 
to  show  that  the  current  and  flux  are  in  time  phase  with  one 
another. 

If  voltage  is  impressed  on  the  motor  at  rest,  current  flows  in 
both  coils  C  and  F.  There  is  a  large  drop  of  voltage  across  F 
since  its  reactance  is  high,  but  only  a  very  small  drop  across  C 
since  its  reactance  is  low,  due  to  the  presence  of  the  short-cir- 
cuited armature  winding  and  thus  at  standstill  a  large  flux 
passes  through  F  and  a  small  flux  through  C. 

The  flux  in  F  is  in  time  phase  with  the  field  current ;  the  cur- 
rent in  the  armature  is  in  phase  opposition  to  the  field  current 
and  therefore  reaches  its  maximum  at  the  same  instant  as  the 
flux  in  F}  and  the  torque  which  is  proportional  to  their  product 
retains  its  sign  as  they  reverse  together. 

When  the  armature  rotates  an  e.m.f.  is  generated  between  the 
brushes  by  the  armature  conductors 
cutting  the  flux  from  F.  This  e.m.f.  is 


at    every   instant    proportional    to    the 
product  of  the  flux  and  the  speed  and  is 
in  phase  with  the  flux  and  is  therefore 
i^1    ^O  degrees  behind  the  e.m.f.  across  F. 
FIG  4go  The  armature  now  acts  at  the  primary 

of  a  transformer  with  the  compensating 

coil  as  secondary  and  it  produces  a  flux  which  transfers  the  speed 
e.m.f.  to  the  compensating  coil  and  the  coil  C  therefore  con- 
sumes a  large  component  of  the  impressed  e.m.f. 
In  Fig.  480, 

/  is  the  line  current  which  flows  in  the  coils  F  and  C. 
3>F  is  the  flux  in  F,  which  is  in  phase  with  the  current. 
EF  is  the  component  of  impressed  e.m.f.  across  the  termi- 
nals of  F. 

Es  is  the  e.m.f .  generated  in  the  armature  by  rotation  and 
transferred  to  the  compensating  coil. 


ALTERNATING-CURRENT  COMMUTATOR  MOTORS  503 

EC  is  the  component  of  impressed  e.m.f.  across  the  termi- 
nals of  C  ;  it  is  equal  to  Es  if  the  coil  C  has  the  same  num- 
ber of  turns  as  the  armature,  and  it  is  in  phase  with  it. 


E  =  ^/E^  +  Ec2  is  the  constant  line  voltage  impressed 
on  the  motor. 

The  e.m.f.  consumed  by  the  impedance  of  the  armature  and  com- 
pensating winding  is  neglected.  As  the  speed  increases  the  e.m.f. 
EC  increases  and  EF  decreases,  the  flux  in  the  main  field  F  de- 
creases and  the  current  and  torque  decrease. 

At  start,  when  the  drop  across  C  is  small,  the  current  is  large 
and  the  main  field  F  is  very  strong.  The  repulsion  motor,  there- 
fore, gives  a  good  starting  torque.  The  field  at  start  will  be 
decreased  to  a  certain  extent  by  the  current  in  the  short-cir- 
cuited coil  undergoing  commutation,  as  in  the  series  motor. 

424.  Commutation.  —  In  the  single-phase  series  motor  and  the 
repulsion  motor  there  are  two  currents  to  be  commutated:  (1) 
the  load  current  and  (2)  the  short-circuit  current  produced  in 
the  coil  under  the  brush  by  the  alternating  flux  of  the  main  field. 

1.  To  reverse  the  load  current  a  m.m.f.  is  required  opposing 
the  m.m.f.  of  armature  reaction  and  strong  enough  to  produce  a 
flux  in  the  opposite  direction  to  the  armature  reaction  flux.     Such 
a  flux  can  be  produced  by  interpoles  placed  between  the  main 
poles  and  excited  by  a  winding  in  series  with  the  main  field  or  it 
can  be  produced  by  a  compensating  winding.     The  conductively 
compensating  winding  is  the  only  one  which  can  give  perfect 
commutation  since  its  m.m.f.  can  be  made  stronger  than  the 
armature  m.m.f.     Commutation  is  assisted  by  the  use  of  high- 
resistance  carbon  brushes. 

2.  To  eliminate  the  short-circuit  current  in  the  coil  under  the 
brush  an  e.m.f.  must  be  generated  in  the  coil  equal  and  opposite 
to  the  e.m.f.  producing  the  short-circuit  current.     The  neutraliz- 
ing e.m.f.  cannot  be  generated  by  the  alternation  of  a  magnetic 
flux  through  the  coil,  since  that  would  require  a  flux  equal  and 
opposite  to  the  field  flux  and  would  destroy  the  field  of  the  motor. 
The  required  e.m.f.  can,  however,  be  generated  by  the  rotation 
of  the  armature  through  a  commutating  field  of  the  proper  inten- 
sity and  position,  but  the  field  must  be  in  quadrature  with  the 
main  field  in  both  time  and  space.     In  the  repulsion  motor  under 
running  conditions  such  a  field  is  produced  in  the  compensating 
coil  C.     The  intensity  of  the  field  varies  with  the  speed  of  the 


504  ELECTRICAL  ENGINEERING 

motor.  Near  synchronous  speed  the  e.m.f.  is  entirely  neutral- 
ized and  the  current  is  wiped  out.  Below  synchronous  speed  the 
current  is  reduced  and  above  synchronous  speed  another  current 
is  produced  as  objectionable  as  before  and  commutation  becomes 
bad  again.  At  standstill  no  neutralizing  e.m.f.  is  produced. 

In  the  single-phase  series  motor  there  is  no  field  in  quadrature 
with  the  main  field  in  time  and  so  no  neutralizing  e.m.f.  can  be 
produced,  but  the  short-circuit  current  is  reduced  by  using  high- 
resistance  preventive  leads  as  explained  in  Art.  420. 

Thus  near  synchronous  speed  the  commutation  of  the  repulsion 
motor  is  better  than  that  of  the  series  motor. 

Since  preventive  leads  are  not  used  in  the  repulsion  motor  the 
short-circuit  current  in  it  at  start  will  be  greater  than  in  the  series 
motor  and  will  weaken  the  main  field  and  decrease  the  starting 
torque. 

While  running  the  short-circuit  current  is  not  so  great,  since  it 
cannot  reach  its  maximum  value  on  account  of  the  self -inductance 
of  the  coil. 

The  repulsion  motor  cannot  be  operated  more  than  40  per  cent, 
above  synchronous  speed  on  account  of  commutation  troubles. 

425.  Compensated  Repulsion  Motor. — The  main  field  wind- 
ing, F,  or  the  torque  field  of  the  repulsion  motor  may  be  omitted 
if  two  extra  brushes  are  placed  on  the  commutator  at  right  angles 
to  the  main  brushes  as  shown  in  Fig.  481.  The  armature  winding 


(«)  (6) 

FIG.  481. — Compensated  repulsion  motor. 

is  thus  made  to  carry  the  exciting  current  and  the  total  reactance 
of  the  motor  is  decreased  and  the  power  factor  improved.  It  is 
not  usually  desirable  to  design  the  armature  with  the  proper 
number  of  turns  for  excitation  and  the  motor  Fig.  481  (a)  may 
be  replaced  by  that  in  (6),  in  which  the  exciting  brushes  are 
supplied  from  taps  on  the  secondary  of  a  transformer  with  its 
primary  in  series  with  the  compensating  coil  C.  With  this 
arrangement  the  speed  may  be  controlled  by  changing  the  tap  to 


ALTERNATING-CURRENT  COMMUTATOR  MOTORS  505 

which  the  exciting  brush  is  connected.  These  motors  have  the 
serious  objection  that  four  sets  of  brushes  are  required  per  pair 
of  poles. 

A  large  number  of  motors,  differing  in  certain  details  from  the 
simple  series  and  repulsion  motors  described  herfe,  have  been  de- 
signed and  have  good  operating  characteristics,  but  a  discussion 
of  them  is  beyond  the  scope  of  this  book. 

426.  Alternating-current  Commutator  Motors  with  Shunt 
Characteristics. — The  single-phase  induction  motor  has  a  speed 
characteristic  similar  to  a  shunt  motor  but  the  speed  cannot  be 
adjusted  and  the  motor  has  no  starting  torque. 

If  the  squirrel-cage  rotor  is  replaced  by  a  direct-current  arma- 
ture with  two  sets  of  short-circuited  brushes  at  right  angles  as 
shown  in  Fig.  482  the  characteristics  of  the  motor  are  not  changed 
but  it  is  possible  to  improve  its  power  factor  and  to  adjust  its 
speed. 

When  the  motor  is  at  rest  there  is  no  e.m.f.  between  the  xx 
brushes  and  no  current  flows  and  therefore  there  is  no  flux  along 
the  xx  axis.  There  is  flux  along  the  yy  axis  and  current  flows 


1C 

(a)  (b)  (c) 

FIG.  482.  FIG.  483. — Single-phase  commutator  motor  with 

power-factor  compensation. 

through  the  yy  brushes.  If  the  motor  is  started  in  either  direc- 
tion the  conductors  cut  the  yy  flux  and  an  e.m.f.  is  generated 
between  the  xx  brushes  in  phase  with  the  yy  flux  and  proportional 
to  the  speed.  This  e.m.f.  causes  a  current  to  flow  in  the  xx 
brushes  which  produces  a  torque  field  along  the  xx  axis  and  thus 
torque  is  developed  as  in  the  single-phase  induction  motor 
(Art.  408). 

There  are  three  ways  of  improving  the  power  factor  of  the 
motor:  (1)  By  introducing  in  the  circuit  of  the  yy  brushes  a 
stator  winding  in  the  xx  axis,  Fig.  483 (a);  (2)  by  introducing  in 
the  circuit  of  the  xx  brushes  a  stator  winding  in  the  yy  axis, 


506 


ELECTRICAL  ENGINEERING 


Fig.  483(6);  and  (3)  introducing  in  the  circuit  of  the  xx  brushes 
an  e.m.f.  in  phase  with  the  line  voltage,  Fig.  483  (c). 

There  are  likewise  three  methods  of  controlling  or  adjusting 
the  speed:  (1)  By  introducing  in  the  circuit  of  the  xx  brushes  a 
stator  winding  in  the  xx  axis,  Fig.  484 (a) ;  (2)  by  introducing  in 
the  circuit  of  the  xx  brushes  a  variable  inductance,  Fig.  484(6); 
and  (3)  by  introducing  in  the  circuit  of  the  yy  brushes  an  e.m.f. 
in  phase  with  the  line  voltage,  Fig.  484 (c). 


Variable  Inductance 

(a)  W  (0} 

FIG.  484. — Single-phase  commutator  motor  with  speed  adjustment. 

427.  Single -phase  Induction  Motor  with  Repulsion  Starting. — 

The  repulsion  motor  in  Fig.  479  (a)  may  be  changed  into  a  single- 
phase  induction  motor  by  short-circuiting  the  commutator. 
When  the  brushes  of  such  a  motor  are  set  at  the  proper  angle  to 
the  stator  winding  a  good  starting  torque  is  produced.  A  centri- 
fugal governor  may  be  set  to  short-circuit  all  the  segments  of 
the  commutator  and  to  raise  the  brushes  when  a  certain  speed 
is  reached  and  the  motor  will  then  operate  as  an  induction 
motor  at  approximately  constant  speed. 


CHAPTER  XIV 
TRANSMISSION  SYSTEMS 

428.  Transmission  Line. — The  transmission  line  carries  the 
electrical  energy  from  the  generating  station  to  the  receiving 
station  or  substation,  where  it  is  either  transformed  into  mechani- 
cal energy  or  distributed  to  the  customers  throughout  the  district. 

The  most  important  characteristics  of  a  transmission  line  are: 
(1)  reliability,  (2)  regulation  and  (3)  efficiency. 

1.  To  insure  reliability  of  service  lines  should,  wherever  pos- 
sible, be  installed  in  duplicate  and  all  the  necessary  protective 
devices  applied. 

2.  For  good  regulation  the  reactance  of  the  line  should  be  as 
small  as  possible  and  therefore  the  frequency  should  be  low.     The 
capacity  of  a  line  draws  a  leading  current,  which  partially  coun- 
teracts the  drop  in  voltage  due  to  reactance  and  so  improves  the 
regulation. 

3.  The  power  losses  in  a  line  are  the  resistance  loss,  which  varies 
as  the  square  of  the  current,  and  the  comparatively  small  losses 
due  to  leakage  over  the  insulators  and  to  the  formation  of  corona 
around  the  conductors. 

To  reduce  the  power  loss  the  resistance  of  the  line  should  be 
made  as  low  as  possible.  This  can  be  done  by  increasing  the 
cross-section  of  the  conductors,  but  the  increased  cost  of  the 
material  required  soon  overcomes  the  saving  due  to  the  increase 
in  efficiency. 

For  a  given  loss  and  a  given  voltage  between  lines  power  can 
be  transmitted  with  a  smaller  amount  of  conducting  material 
three-phase  than  either  single-phase  or  two-phase. 

429.  Relative  Amounts  of  Conducting  Material  for  Single-, 
Two-  and  Three-phase  Transmission  Lines. 

Let       P  =  power  input  to  the  line  in  watts. 

p  =  per  cent,  loss  of  power  in  the  line  resistance  due  to 

full-load  current. 
/  =  full-load  current. 

507 


508  ELECTRICAL  ENGINEERING 

cos  B  =  power  factor. 

r  =  resistance  of  each  conductor. 
n  —  number  of  conductors  in  the  system. 
The  loss  in  the  line  is 


and  the  resistance  of  each  conductor  is 


100  rc/2 

For  the  same  voltage  E  between  conductors  the  current  is 
P 


E  cos  B 

P 


,  single-phase, 
-,  two-phase, 


2E  cos 
p 

I  =  — 7= —      — ,  three-phase, 
V3E  cos  0 

and  the  resistance  per  conductor  is 

pP  X  E2  cos2  6       .  nn_  pE2  cos20     . 
r  =        ICQ  x  2P2       =  ~T '  single-phase, 

pP  X  4#2  cos2  0  ,  p#2  cos2  e   . 

100  X  4P2  ~P '  two-phase, 


pP  X  3#2  cos2  S      n  m  p#2  cos20   .,          , 
r  =         100  X  3  P2  ~P '  three-phase. 

Since  the  single-phase  line  has  only  two  conductors  while  the 
two-phase  line  has  four  the  amount  of  copper  required  for  both  is 
the  same.  The  three-phase  line  consists  of  three  conductors  of 
the  same  section  as  the  two-phase  conductors  and,  therefore,  the 
amount  of  copper  required  for  a  three-phase  line  is  only  75  per 
cent,  of  that  required  for  a  two-phase  or  single-phase  line  with  the 
same  per  cent,  power  loss  and  the  same  maximum  voltage  between 
lines. 

430.  Reactance. — The  inductance  of  a  line  per  mile  of  con- 
ductor is  (equation  163) 

74  logic  p  +  0.0805  Y  10~3  henrys 

where  D  is  the  distance  between  conductors, 

and  R  is  the  radius  of  the  conductors. 

The  reactance  of  the  line  per  mile  of  conductor  is 
X  =  27T/L  ohms. 


L  =  (o 


TRANSMISSION  SYSTEMS  509 

The  reactance  could  be  decreased  by  decreasing  the  distance 
between  the  conductors  or  by  increasing  the  radius,  but  these 
quantities  are  fixed  by  other  considerations  than  the  reactance 
and  reactance  drop. 

431.  Capacity.  —  The  capacity  of  a  line  per  mile  of  conductor 
between  the  conductor  and  neutral  is 

00   O 

C  =  -  —  10-9  farads  (equation  38). 

,        D  —  ri 
logio  —  ft— 

This  value  applies  for  each  conductor  of  a  single-phase  or  poly- 
phase line.  If  the  conductors  of  a  three-phase  line  are  suspended 
in  one  plane  instead  of  in  the  form  of  an  equilateral  triangle  the 
capacity  of  the  central  conductor  is  slightly  greater  than  that  of 
the  others,  but  since  all  lines  are  transposed  the  total  capacity  of 
each  of  the  three  is  the  same  and  is  given  with  sufficient  accuracy 
by  the  formula  above  if  the  distance  D  is  taken  as  the  shortest 
distance  between  conductors. 

The  capacity  reactance  per  mile  of  conductor  is 


and  the  charging  current  per  mile  of  conductor  is 


where  e  is  the  voltage  between  conductor  and  neutral. 

For  transmission  lines  up  to  50,000  volts  the  capacity  is  very 
small  and  its  effect  on  the  regulation  may  be  neglected.  If,  how- 
ever, any  part  of  the  transmission  is  carried  out  through  under- 
ground cables,  the  capacity  may  be  very  largely  increased  and 
may  not  be  negligible.  Above  50,000  volts  the  capacity  of  the 
line  must  be  considered  in  calculating  the  regulation.  For  lines 
up  to  100  miles  in  length  and  for  voltages  up  to  100,000  volts  the 
capacity  of  each  conductor  may  be  considered  as  a  condenser 
connected  at  the  center  of  the  line  between  conductor  and  neutral. 
If  more  accurate  results  are  necessary  the  fact  that  both  the  re- 
actance and  the  capacity  of  the  line  are  distributed  over  the  whole 
length  must  be  taken  into  account. 

Due  to  the  presence  of  the  charging  current  in  a  line  the  current 
flowing  into  the  receiving  circuit  may  be  very  much  larger  than 
the  current  entering  the  line  at  the  generating  station. 


510 


ELECTRICAL  ENGINEERING 


432.  Voltage  and  Frequency.— Voltages  up  to  150,000  volts 
are  now  in  use  for  the  transmission  of  large  amounts  of  power 
over  long  distances.  The  voltage  employed  in  any  given  system 
is  as  a  general  rule  approximately  one  thousand  times  the  length 
of  the  line  in  miles. 

Power  is  usually  generated  and  transmitted  at  either  25  cycles 
or  60  cycles.  With  25  cycles  the  reactance  drop  in  the  line  is  less 


Distance  between  Lines  -  Incbe 

o  8  £  §  S  §  g 

^ 

^ 

^ 

^ 

"^ 

^ 

S* 

' 

' 

,  " 

-^ 

)      10     20     30     40      50     00     70     80     90     100  11 
Kilovolts  between  Lines 

FIG.  485. — Spacing  of  conductors. 


than  with  60  cycles  and  therefore  the  voltage  regulation  is  better. 
In  the  case  of  very  long  high-voltage  lines  the  increased  charging 
current  at  the  higher  frequency  may  counteract  the  larger  react- 
ance drop  of  voltage.  Where  power  is  required  for  lighting  60 
cycles  is  necessary  unless  frequency  chargers  are  installed. 

433.  Spacing  of  Conductors. — The  distance  between  the  con- 
ductors of  a  transmission  line  depends  both  on  the  voltage  and 
also  on  certain  points  in  the  mechanical  design,  such  as  the  mate- 


Eo 


FIG.  486. — Single-phase  transmission  line. 


rial  of  the  conductor,  length  of  span  and  the  amount  of  sag  al- 
lowed. The  curve  in  Fig.  485  gives  approximately  the  relation 
between  the  spacing  of  the  conductors  and  the  voltage. 

434.  Single-phase  Transmission  Line. — 1.  A  single-phase  trans- 
mission line,  Fig.  486,  delivers  5,100  kw.  to  a  receiver  circuit  at 


TRANSMISSION  SYSTEMS  511 

60,000  volts.     If  the  power  factor  of  the  load  is  85  per  cent., 
find  the  generator  voltage. 

r  =  resistance  of  the  line  =  20  ohms. 
x  =  reactance  of  the  line  =  50  ohms. 

The  power  delivered  to  the  receiver  circuit  is 

P  =  El  cos  <f>  =  5,100,000  watts, 
where  E  =  60,000  is  the  receiver  voltage 

and  cos  0  =  0.85  is  the  power  factor; 

the  current  is  therefore 

P  5,100,000 

"  E  cos  <f>  ~~  60,000  X  0.85   " 

The  vector  diagram  is  drawn  with  the  current  01  =  I  as  hori- 
zontal. 

The  receiver  e.m.f.  OE  =  E  leads  the  current  by  -an  angle  <j> 
and  has  two  components 

OEi  =  EI  =  E  cos  <£  in  phase  with  /  and 
OE2  =  E2  =  E  sin  0  in  quadrature  ahead  of  /. 
The  voltage  consumed  in  the  resistance  of  the  line  is  Ir  in 
phase  with  7;  the  voltage  consumed  in  the  reactance  of  the  line 
is  Ix  in  quadrature  ahead  of  /. 

The  component  of  the  generator  e.m.f.  in  phase  with  I  is 

El  +  Ir  =  E  cos  0  +  Ir 
and  the  component  in  quadrature  ahead  of  /  is 

E2  +  Ix  =  E  sin  <t>  +  Ix, 
and  therefore  the  generator  e.m.f.  is 

EG  =  \/(E  cos 0  +  Ir)2  +  (E  sin  0  +  Ix)'2, 
or  substituting  the  numerical  values 

EG  =  V(60,000  X  0.85  +  100  X  20)2+(60,000  X  0.52+100X50)2 

=  64,000  volts. 
The  e.m.f.  consumed  in  the  line  is 

iVr2  +  x2  =  100  V202  +  502  =  5,400  volts. 
The  loss  of  power  in  the  line  is 

72r  =  1002  X  202  =  200,000  watts. 

-  200  kw. 

The  power  factor  at  the  generator  is 
cos  to  *  l^+Jr  =  53,000  =  Q  ^  =  ^ 

£JQ  D4,UUU 


512  ELECTRICAL  ENGINEERING 

Using  rectangular  coordinates  and  taking  the  current  as  axis, 
the  e.m.f.  at  the  receiver  terminals  is 

E  =  E cos  0  +  jE  sin  </>, 
the  e.m.f.  consumed  in  the  impedance  of  the  line  is 

E'  =  IT  +  jlx, 
and  the  generator  e.m.f.  is 

EG  =  E  +  E'  =  (E  cos  0  +  Ir)  +  j(E  sin  4>  +  Ix), 
and  its  absolute  value  is 

EG  =  \/(E  cos  0  +  Ir)2  +  (E  sin0  +  Ix)2. 

The  capacity  of  the  line  has  been  neglected  in  this  example. 

2.  A  transmission  line  of  impedance  Z  =  r  +  jx  delivers  power 
to  a  receiver  circuit  of  admittance  F  =  g  —  jb  at  a  constant 
voltage  E.  If  the  capacity  of  the  line  is  assumed  to  be  concen- 
trated at  the  center  determine  the  charging  current  of  the  line, 
the  total  current  delivered  by  the  generator  and  the  terminal 
voltage  of  the  generator. 

The  condensive  reactance  of  the  line  is 


4    '«  j-fir 

^2 

Js 

< 

E           < 

\ 

—  vvW\  — 
1 

<b  — 

—  r  00000 

\  __.  VWV*  —  "^ 

FIG.  487.  —  Single-phase  transmission  line  with  capacity. 

where  /  is  the  frequency  of  the  impressed  e.m.f.  and  C  is  the  ca- 
pacity of  the  line  in  farads;  the  condensive  susceptance  of  the 
line  is  j 

yc  =  bc  =  - 

O/c 

and  it  is  represented  as  a  condenser  connected  across  the  line 
(Fig.  487). 

The  current  in  the  receiver  circuit  is 


and  the  e.m.f.  at  the  center  of  the  line  is 


TRANSMISSION  SYSTEMS  513 

The  charging  current  of  the  line  is 

(g    -  jb)       ,  (451) 


and  the  current  from  the  generator  is 


-  J6+J5.1  +-(g-  jb)-   (452) 


The  terminal  e.m.f.  of  the  generator  is 


•j,  (r  +  jz)*(g  -  jb) 


(r  +  jx)  (g  -  jb  +  j  D  +  j  \  (r  +  jxY(g  -  jb) } 


=  E        ......      .      ,,„..._.. 

(453) 

For  lines  of  small  capacity  the  last  term  may  be  neglected  and 
equation  (453)  reduces  to 


1  +  (r+jx)    g-jb+j-  (454) 

This  is  equivalent  to  replacing  the  capacity  susceptance  bc 
at  the  center  of  the  line  by  a  condenser  of  susceptance  -^  at  the 

A 

receiver  end  of  the  line. 

3.  A  single-phase  transmission  line  delivers  10,000  kw.  at 
100,000  volts  to  a  receiver  circuit  of  85  per  cent,  power  factor; 
find  the  voltage,  currenlLand  power  factor  at  the  generating  end 
of  the  line,  the  impedance  drop  and  power  loss  in  the  line  and 
the  charging  current.  Find  also  the  generator  voltage  required 
to  give  a  receiver  voltage  of  100,000  volts  at  noToadT 

Length  of  line  =  100  miles. 
Size  of  wire  =  No.  000  B.  &  S.  copper. 
Diameter  of  wire  =  2R  =  0.41  in. 
Distance  between  wires  =  D  =  100  in. 
Frequency  =  60  cycles  per  second. 

The  inductance  or  coefficient  of  self-induction  of  each  wire  of 
the  line  is,  by  equation  (163), 

Li  =  /0.74  logio/o  +  0.0805)  10~3  henrys  per  mile, 

33 


514  ELECTRICAL  ENGINEERING 

and  therefore  the  inductance  of  the  line  consisting  of  two  wires  is 
L  =  200Li  =  200  (0.74  logio  Q~;  +  0.0805)  10~3=  0.41  henrys; 

the  inductive  reactance  of  the  line  is 
x  =  27T/L 

=  2  X  3.14  X  60  X  0.41  =  154.5  ohms. 

The  capacity  of  each  wire  to  neutral  is,  by  equation  (38), 

38  8 

Ci  =  -     —  pj  -  5  10~9  farads  per  mile, 
,          LJ  —  rC 


and  the  capacity  between  wires  is 


2       ,        D  - 
logio  — n~ 


"8  farads  per  mile  of  line; 


104. 

c  =  iooc2  =  100  x  -        T   io~9 


therefore,  the  capacity  of  the  line  is 

1  t/  t/«w 

Iogl°02l5 
=  0.73  10~6  farads. 

The  condensive  reactance  is 

Xc  =  2irfC 

=  2  X  3.14  X  60XX  0.73  X  10-  =  364°  °hmS' 
and  the  condensive  susceptance  of  the  line  is 


The  resistance  of  the  line  at  20°C.  is 
I 


r  =  p 


dr.  mils 


A  200  X  5,280 

(410)2         =65.4  ohms. 

The  load  delivered  to  the  receiver  is 

P  =  El  cos  <j)  =  10,000,000  watts, 
but  E  =  100,000  and  cos  <j>  =  0.85; 

therefore,  the  current  in  the  receiver  circuit  is 

P  10,000,000 

=  E  cos  0  ~  100,000  X  0.85  ~  L1/  amp'; 


TRANSMISSION  SYSTEMS  515 

the  power  component  is 

I  P=  /cos  0  =  117  X  0.85  =  100  amp.; 
the  quadrature  component  is 

Iw  =  I  sin  0  =  V/2  -  Ip2  =  A/1172  -  1002  =  60  amp. 
The  admittance  of  the  receiver  is 

1          H7 
~  E~  100,000  ~ 
the  conductance  is 

/  cos  <t>          100 


and  the  susceptance  is 


7  sin  0  60 

"-    =  = 


The  generator  voltage  is,  by  equation  (453), 
E0  =  E  j  1  +  (r  +  jx)  (g  -  jb  +  j  |)  +  j  |(r  +  jx)*(g  -  jb)  ) 
and  substituting  the  values  obtained  above 

+  (65.4  +  154.5J)  (O.OOI  -  0.0006J  + 


0  00097^ 
+  -  —      -  j  (65.4  +  154.5J)  2  (0.001  -  0.0006J) 


and  simplifying 

EQ  =  ^(1.137 

and  the  absolute  value  is 


EQ  =  #(1.137)2+  (0.114)2 

=  100,000  X  1.142  =  114,200  volts. 
The  current  from  the  generator  is,  by  equation  (542), 

lo  =  E  [g-jb+jb.  {  1  +  r-^p-  (g-jb) 
=  &  [o.OOl  -  0.0006?  +  0.000275.? 


=  £^(0.000984  -  0.  00029;'), 
and  its  absolute  value  is 


IQ  =  100,000  V(0.000984)2  +  (0.00029)2  =  102.6  amp. 


516  ELECTRICAL  ENGINEERING 

The  charging  current  of  the  line  is,  by  equation  (541), 
Ic=jbcE  [1  +  !+&(,_#)) 

=  E  (-  0.0000158  +  0.00031J), 
and  its  absolute  value  is 

Ic  =  100,000  V(0.0000158)2+  (0.00031)2  =  31  amp. 
=  26.5  per  cent,  of  the  receiver  current. 

At  no  load  E  =  100,000,  g  =  0,  b  =  0  and  the  generator  vol- 
tage is 


and  its  absolute  value  is 


EQ  =  1  00,000  V(a9786)2-{-  (0.0098)  2  =  97,860  volts. 
The  power  factor  at  the  generator  may  be  found  by  reference 
to  the  diagram  in  Fig.  488. 

EG 


E0  leads  E  by  an  angle  <£',  where 


.  /  JL  _L  .^  4-\    -*  f\ 

tan  ,'  ,.  -  =  0.10; 


IG  lags  behind  E  by  an  angle  <£",  where 

tan-0"  =  ~j^  =  0.0296; 

and  IG  lags  behind  EG  by  an  angle  <bG  =  $'  + 
tan  ,e  =  tan      '  +     '     = 


and  0G  =  7°307; 

the  power  factor  at  the  generator  is 

cos  4>G  =  cos  7°  30'  =  0.99 

=  99  per  cent. 

The  impedance  drop  in  the  line  is  found  very  approximately  as 
Ez  = 


=  117V(65.4)2+  (154.5)2  =  19,650  volts 
=  19.6  per  cent,  of  the  receiver  voltage. 


TRANSMISSION  SYSTEMS 


517 


The  power  loss  in  the  line  is  found  approximately  as 

T  T 

Pr    =   Ir2  ~   +  I'2  - 
°     2  2 

=  (102.6)  234  +  (117)234 

=  826,000  watts 

=  826  kw. 

=  4.13  per  cent,  of  the  output. 

435.  Three-phase  Transmission  Line.  —  A  three-phase,  trans- 
mission line  delivers  30,000  kva.  at  100,000  volts,  60  cycles  to  a 
receiving  circuit  of  88  per  cent,  power  factor;  determine  the 
voltage  and  current  and  power  factor  at  the  generating  station, 
the  charging  current  and  charging  kilovolt-amperes  of  the  line  and 
the  efficiency  of  the  transmission.  Determine  also  the  rise  in 
voltage  at  the  terminals  of  the  receiving  circuit  if  full  load  is 
suddenly  removed. 

Length  of  line  100  miles. 

Conductor  =  No.  0000  B.  &  S.  copper  of  97  per  cent,  conduc- 
tivity. 

Diameter  of  conductor  =  2R  =  0.46  in. 

Distance  between  conductors  =  100  in. 


FIG.  489.  —  Three-phase  transmission  line. 

In  Fig.  489 

E  =  voltage  between  lines  at  the  receiving  end  =  100,000. 

•I  rvr\  000 

e  =  voltage  between  lines  and  neutral  =  --  W-  =  57,600. 

\/6 


E0  =  voltage  between  lines  at  the  generating  station. 
e0  =  voltage  between  lines  and  neutral  = 


ET 


. 

V3 

I  =  load  current  in  each  conductor. 

Ic  =  charging  current  in  each  conductor. 

IG  =  current  per  conductor  at  the  generating  station. 


518  ELECTRICAL  ENGINEERING 

Y  =  \/V  +  b2  =  admittance  of  each  phase  of  the  receiving 

circuit. 
Z  —  \/r2  -j-  xz  =  impedance  of  each  line  including  the  step- 

up  and  step-down  transformers. 
bc  =  capacity  susceptance  of  each  conductor  to  neutral,  which 

is  assumed  to  be  concentrated  at  the  center  of  the  line. 

Resistance  of  a  single  conductor  of  97  per  cent,  conductivity 
at  20°C.,  100  miles  long  =  r'  =  0.2667  X  100  =  26.67  ohms 
(article  86). 

30,000,000 

=       x  100,000  =  173  amp' 


Resistance   drop   in   each  conductor  =  Ir'  =  173  X  26.67  = 
4,620  volts  =     '  10Q  QOQ  •    =  8  per  cent. 


Assuming  a  resistance  drop  in  step-up  and  step-down  trans- 
formers of  0.5  per  cent,  each,  the  total  resistance  drop  per  line  is 
8  +  2(0.5)  =  9  per  cent. 

g 

r  =  equivalent    resistance     of    each   line  =  26.67  X  5  =  30 

o 

ohms. 

Inductance  of  each  conductor  is 


L  =  (0.74  log™          +  0.0805)10-  3X  100  =  0.203  henry. 

O.Zo 

Inductive   reactance   of  each   conductor  =  xr  =  2irfL  =  2  X 
3.14  X  60  X  0.203  =  77  ohms. 

Reactance    drop    per    conductor  =  lx'  =  173  X  77  =  13,321 

13,321  X  100       00 
V°lts=         100000"     =  23.1  per  cent. 


Reactance  drop  in  each  transformer  =  3.45  percent,  (assumed). 
Total  reactance  drop  per  line  =  23.1  +  2(3.45)  =  30  per  cent. 

77  X  30 
x  =  equivalent  reactance  of  each  line  =  —^-^  —  =  100  ohms. 

Zo.  1 

Capacity  of  each  conductor  to  neutral  is 

00    O 

C  =  -  -A-OQ  X  10~9  X  10°  =  !-475  X  10~6  farads- 

i  1UU  —  (J.AO 

logl°  ~~ 


TRANSMISSION  SYSTEMS  519 

Capacity    reactance    of    each    conductor    to  .neutral  =  xc  = 
106 


=  1,800  ohms. 


2  X  3.14  X  60  X  1.475 

bc  =  capacity  susceptance  per  line  =  —  =  0.000555. 

xc 

T  17S 

Y  =  load  admittance  per  phase  =  -  =    IQQ  QQQ   =  0-003. 


cos  0  =  load  power  factor  =  0.88. 

g  =  load    conductance    per    phase  =-  -         -  =  Y  cos  0  = 

6 

0.003  X  0.88  =  0.00264. 

7  sin  6       .,   . 
b  =  load  susceptance  per  phase  =  -  -  =  Y  sin  9  = 

6 

0.003  X  0.475  =  0.001425. 

Voltage  between  lines  and  neutral  at  the  generating  station  at 
full  load  is 

ft*  =  6      1  +  (r  +  jx)(g  -  jb  +j  equation  (454). 


=  e[l  +  (30+100.?)  1  0.00264  -j  (0.001425  -0.0002775)  J 


=  e(1.1942  +  0.23;) 
and  its  absolute  value  is 


e0  =  D2    +  023    -  70,000  volts 

V3 

0  230 
and  it  leads  the  terminal  e.m.f.  e  by  angle  6'  =  tan"1 


tan-1  0.1925  =  10.9  deg. 

EG  =  voltage  between  lines  =  \/3eG=  \/3X  70,000=  121,600  volts. 
The  current  per  conductor  at  the  generating  station  is 

J0  = 


=  e[o.00264-0.001425j  +  0.000555J  X 

j  i  +  SO^IOQ,  ^  OQ264  _  0.001425j)  j  ] 


=  e  [o.00258  -  O.OOOSlj]  . 
and  its  absolute  value  is 

IG  =  -  -W-  Jo.002582  +  0.000812  =  161  amp. 


520  ELECTRICAL  ENGINEERING 

0  00081 

It  lags  behind  the  terminal  e.m.f.  e  by  angle  0"  =  tan-1  •f^^KO 

O.OOZoo 


=  tan~!  0.314  =  17.45  degrees. 

I0  lags  behind  eG  by  angle  0G  =  d'  +  0"  =  10.9  +  17.45  = 
28.35  degrees  and  the  power  factor  at  the  generator  is  cos  00  = 
cos  28.35  degrees  =  0.88  =  88  per  cent. 

The  charging  current  per  line  may  be  taken  as  ijie  product  of 
the  capacity  susceptance  bc  and  the  average  of  the  two  voltages 
e0  and  e,  thus, 
7c  =  «  +  •  bc  ,  70,000  +  57,600  x  Q  Q^ 

=  63,800  X  0.000555  =  35.4  amp. 
and  the  charging  kilo  volt-amperes  for  the  system 


3  X  63,800  X  35.4        ,„_, 

=  6>800  kva- 


The  copper  losses  in  the  three  lines  and  the  step-up  and  step- 
down  transformers  may  be  taken  as  3(/G2  +  I2)  „   =   3(16  I2  + 


O 

1732)15  =  2,520,000  watts  =  2,520  kw.  =  0>88  X  100 


=  9.5  per  cent. 

Assuming  the  iron  losses  in  the  transformers  to  total  1  per 
cent,  the  total  loss  is  10.5  per  cent,  and  the  efficiency  of  the 
transmission  from  the  generator  busbars  to  the  low-voltage  bus- 
bars in  the  receiving  station  is  89.5  per  cent. 

If  the  load  is  suddenly  removed,  the  rise  in  the  receiver  voltage 
may  be  found  from  the  following  relations  : 

g  =  0,  b  =  0,  eG  is  assumed  to  be  maintained  constant,  and 

e0  =  e  jl  +  (r+jx)  (j|)  }  =  e{l  +  (30  +  100J)  (0.0002775J)  1 

=  6(0.97225  -O.OOSj) 
and  taking  absolute  values 

70,000  =  e0  =  e\/  (0.97225)  2+  (0.008)  2  =  e  X  0.9723 


0^23  -  =  72>°00 


and  E  =  \/3e  =  \/3  X  72,000  =  124,700  volts. 


The  regulation  is        >        ~J  [QQ>000  X  100  per  cent.  =  24.7  per 
cent. 


TRANSMISSION  SYSTEMS  521 

436.  Application  of  a  Synchronous  Phase  Modifier  to  a  Trans- 
mission System.  —  In  the  transmission  system  worked  out  in  the 
last  article,  the  regulation  is  poor  and  the  efficiency  is  low.  These 
can  both  be  improved  by  installing  in  the  receiving  station  a 
synchronous  machine  to  operate  as  a  phase  modifier.  It  should 
be  over-excited  and  draw  full  leading  kilovolt-amperes  at  periods 
of  full  load  to  keep  the  receiver  voltage  up  and  should  be  under- 
excited  and  draw  lagging  kilovolt-amperes  at  periods  of  light  load 
to  keep  the  voltage  down.  With  such  an  equipment  the  voltage 
at  the  generating  station  may  be  maintained  at  a  suitable  con- 
stant value  and  the  receiver  voltage  may  likewise  be  maintained 
constant  at  some  lower  value  by  means  of  an  automatic  vol- 
tage regulator  excited  from  the  constant  receiver  voltage  and 
operating  on  a  shunt  to  the  field  rheostat  of  the  phase  modifier. 

The  equation  eG  =  e  j  1+  (r-\-  jx)  (g  —  jb  +  j  ~)  |was  derived 

on  the  assumption  that  the  capacity  susceptance  was  connected 
at  the  center  of  the  line,  but  the  same  result  is  obtained  if  one-half 
of  this  susceptance  is  connected  at  the  receiving  end.  When  the 
synchronous  phase  modifier  is  added,  Fig.  490  (a),  the  equation 
becomes 


e0 


=  e  [l  +  (r  +  jx)  (g  -  jb  +  j  ^  ±  j&.)  }  (455) 


where  bs  is  the  susceptance  per  phase  of  the  phase  modifier; 
bs  is  positive  when  drawing  a  leading  current  and  negative  when 
drawing  a  lagging  current. 

The  rating  of  the  phase  modifier  required  for  any  system  can 
best  be  found  by  a  graphical  construction  and  equation  (455) 
used  only  as  a  check. 

1.  If  it  is  desired  to  operate  the  system  of  Art.  435  with  a  con- 
stant voltage  E  =  100,000  volts  and  a  constant  generating  sta- 
tion voltage  EG  =  110,000  volts,  the  rating  of  the  required  phase 
modifier  may  be  found  as  follows  : 

In  Fig.  490 

Oa  =  receiver  voltage  to  neutral  =  e  =  --  h=—  =  57,600  volts. 


ab  =  receiver  current  per  phase  =  I  =  173  amp.  at  full  load. 
Cos  6  —  load  power  factor  =  0.88. 

ac  =  resistance  drop  per  line  due  to  the  load  current  =  Ir  = 
173  X  30  =  5,200  volts, 


522 


ELECTRICAL  ENGINEERING 


cd  =  reactance  drop  per  line  due  to  the  load  current  =  Ix  = 
173  X  100  =  17,300  volts.  _ 

ad  =  impedance  drop  per  line  =  /  \/r2  +  z2  =  173-\/302 


=  173  X  104.4  =  18,060  volts  leading  the  current  by  angle  a  = 
x  100 


/  *J\J 

Assuming  the  line  capacity  to  be  represented  by  the  susceptance 
^  at  the  receiver,  Fig.  490  (a),  the  charging  current  flowing  over 

the  line  is  e  X  ^  =  57,600  X  0.0002775  =  16  amp.  =  ah. 


Radius  =eG  =63300 


FIG.  491. 


ak  =  impedance  drop  due  to  current  ah  =  16^302  +  1002  = 
1,670  volts.     If  dg  is  drawn  through  d  making  angle  a  =  tan-1 

with  the  charging  current  line  and  dl  is  cut  off  equal  to  ak, 

then  01  is  the  voltage  to  neutral  required  at  the  generating  sta- 
tion, without  the  phase  modifier  =  70,000  volts  from  the  dia- 

gram.    This  value  was  also  obtained  in  Art.  435. 

i         1  1  n  nnn 


About  0  draw  a  circle  of  radius  eG  = 


=  63,360 


= 

to  cut  dg  at  /;  then  the  line  If  represents  the  impedance  drop  per 
line  produced  by  the  leading  current  flowing  to  the  phase  modifier. 


TRANSMISSION  SYSTEMS  523 

Since  ad  is  the  impedance  drop  due  to  the  load  current  the  line 
If  will  represent  the  kilovolt-ampere  rating  of  the  phase  modifier 
to  the  same  scale  that  the  line  ad  represents  the  load  kilovolt- 
amperes  of  30,000.  The  rating  of  the  phase  modifier  is  therefore 

-^  X  30,000  =  12,500  kilovolt-amperes. 
ad 

At  no  load  the  voltage  required  at  the  generating  station, 
without  the  phase  modifier,  is  represented  by  the  line  Ok  but 
since  eG  is  maintained  constant  at  63,360  volts,  the  phase  modifier 
must  be  under-excited  and  draw  a  lagging  current  to  cause  an 
impedance  drop  represented  by  kf,  where  /'is  the  point  of  inter- 
section of  the  eG  circle  with  the  line  kaf  making  angle  a  with 
ag.  kf  is  approximately  equal  to  If  and  the  phase  modifier  will 
be  carrying  full  lagging  current. 

It  is  not  usual  to  maintain  constant  voltage  over  such  a  wide 
range  and  the  rating  of  the  phase  modifier  may  be  reduced  con- 
siderably especially  if  a  greater  difference  between  eG  and  e  is 
allowable. 

2.  If  it  is  required  to  maintain  a  constant  receiver  voltage,  E 
=  100,000  from  full  load  of  30,000  kva.  at  88  per  cent,  power 
factor  to  one-fourth  load  of  7,500  kva.  at  75  per  cent,  power 
factor,  determine  the  rating  of  the  phase  modifier  and  the  most 
suitable  voltage  at  the  generating  station. 

The  point  li  in  Fig.  491,  corresponds  to  point  I  in  Fig.  490  and 
is  found  in  the  same  way;  and  angle  dl  =  Q  =  cos"1  0.88. 

At  one-fourth  load  of  7,500  kva.  at  75  per  cent,  power  factor. 

a&2  =  —r-  =  —r-  =  43.25  amp.  =  load  current. 
Cos  62  =  0.75  =  load  power  factor. 

IT 

Angle  d2ab2  =  angle  diabi  =  a  =  tan"1  -• 

Points  d2  and  Z2  are  found  in  the  same  way  as  d  and  I,  Fig.  490. 

To  enable  the  phase  modifier  to  be  used  to  its  full  capacity 
leading  at  full  load  and  full  capacity  lagging  at  one-fourth  load, 
the  generator  voltage  must  have  the  value  represented  by  the 
radius  of  a  circle  drawn  about  0,  which  will  make  the  two  inter- 
cepts /]Zi  and  /2?2  equal.  The  required  value  of  eG  is  65,350  volts 
and  E0  =  \/3  X  65,350  =  113,000  volts.  The  capacity  of  the 

phase  modifier  is  ~r  X  30,000  =  9,100  kva.   and  its    suscept- 


ance  per  phase  at  full  load  is  ba  =  >         =  0.00091. 


524  ELECTRICAL  ENGINEERING 

Referring  to  Fig.  490  (a)  the  current  in  each  line  at  full  load  is 

f  .  bc   ,     ,    } 

IG  =  e  I  g  —  jb  H-j  ^  +  A  j 

=  6(0.00264  -  j(0.001425  -  0.0002775  -  0.00091)) 
=  e  (0.00264  -  0.00025J) 


-2 


and          IG  =  57,600  \0.0264   +  0.00025      =  153  amp. 

The    full-load    copper    loss  =  3  X  (153)2  X  30  =  2,115,000 

2115 
watts  =  2,115  kw.  =  30)00(^  x  Q88X  100  =  8  per  cent. 

Taking  the  transformer  iron  losses  as  1  per  cent.,  the  efficiency 
of  the  transmission  is  91  per  cent.  The  losses  in  the  phase 
modifier  should  be  subtracted. 

437.  High-voltage  Direct-current  System. — The  transmission 
of  power  over  long  distances  by  direct  current  is  limited  by  the 
difficulty  of  obtaining  a  sufficiently  high  voltage.  In  the  Thury 
system  the  required  line  voltage  is  obtained  by  connecting  a 
number  of  series-wound  generators  in  series  and  the  line  current 
is  automatically  maintained  constant.  Series  machines  are 
built  generating  from  1,500  to  5,000  volts  with  a  single  commuta- 
tor. A  single  generating  unit  may  have  two  commutators  and 
two  units  may  be  driven  by  one  prime  mover  giving  a  direct 
voltage  up  to  20,000  volts  per  set.  If  five  such  machines  are 
connected  in  series  the  voltage  between  lines  will  be  100,000  volts. 

To  maintain  constant  line  current  with  variable  load,  the  gen- 
erator voltage  must  be  varied;  this  is  accomplished  by  means  of 
an  automatic  regulator  which  moves  the  brushes  and  varies  the 
resistance  in  a  rheostat  or  diverter  shunting  part  of  the  field 
winding.  In  some  cases  the  speed  of  the  prime  mover  is  varied. 
The  regulator  may  be  set  for  any  required  current.  In  the  sys- 
tems at  present  in  operation  currents  from  50  to  200  or  300  amp. 
are  employed. 

It  is  impossible  to  insulate  rotating  machines  for  the  high 
voltages  required  for  transmission  and  it  is  therefore  necessary 
to  mount  the  generators  on  insulated  platforms  and  to  couple 
them  to  the  prime  movers  by  insulating  couplings. 

The  various  generating  units  need  not  be  included  in  a  single 
station  but  may  be  located  at  any  point  along  the  proposed  line. 
In  this  way  a  number  of  small  generating  stations  may  be  linked 
up  into  one  large  system  and  since  the  current  is  constant  they 
cannot  be  overloaded  under  any  condition. 


TRANSMISSION  SYSTEMS 


525 


Fig.  492  shows  the  layout  of  a  60, 000- volt,  150-amp.  system. 

At  periods  of  light  load  any  generator  may  be  taken  out  of 
service  by  disconnecting  the  regulator  and  moving  the  brushes 
over  to  the  position  of  zero  voltage  and  then  closing  the  short- 
circuiting  switch.  To  put  it  in  service  again  it  is  brought  up  to 
speed,  the  short-circuiting  switch  is  opened  and  the  regulator  set 
for  the  required  current. 

The  substation  equipment  is  very  similar  to  that  in  the  gen- 
erating station  and  like  it  may  be  located  in  one  or  more  stations. 
If  it  is  necessary  to  operate  the  motors  at  constant  speed  a  regu- 
lator is  provided  which  maintains  the  speed  constant  by  moving 
the  brushes  and  shunting  part  of  the  motor  field  current. 


Q  \r    Short  Circuiting 
Bwitch 


Generator  Out  of  Service 

(a) 


10000  Volts         8000  Volts          2000  Volts 

150  Amperes       150  Amperes       150  Amperes        160  Ampei 

Generating  Station 


150  Amperea 

"  f 


Or  Stations 


20000  Volte 
150  Amperes 


VR 
L@jaJ 

FIG.  492. — Thury  or  series  system. 


The  motors  must  drive  through  insulating  couplings  and  are 
put  in  operation  by  opening  a  short-circuiting  switch  and  moving 
the  brushes  to  the  proper  position.  The  regulator  is  then  con- 
nected in  and  takes  care  of  the  speed. 

In  the  majority  of  terminal  stations  the  power  which  has  been 
transmitted  by  direct  current  is  converted  into  alternating  cur- 
rent for  local  distribution,  the  series  motors  driving  alternat- 
ing-current generators. 

438.  Advantages  and  Disadvantages  of  the  Thury  or  Series 
System. — Advantages. — 1.  Power  factor  is  always  unity  as  there 
is  no  reactance  drop.  Cables  carrying  direct  currents  can  be 
laid  in  iron  pipes  if  necessary. 

2.  Higher  voltages  may  be  used  for  the  same  line  insulation. 
Direct  voltages  may  be  double  the  effective  alternating  voltages 
since  there  are  no  dielectric  losses  in  the  cables. 

3.  Two  conductors  only  have  to  be  insulated  and  the  center 
point  of  the  system  may  be  grounded  to  reduce  stresses. 


526  ELECTRICAL  ENGINEERING 

4.  A  number  of  stations  can  be  operated  in  series  and  a  new 
station  may  be  connected  to  the  line  at  any  point.     The  indi- 
vidual stations  are  entirely  independent  of  local  overloads  or  lack 
of  demand  since  the  current  is  maintained  constant  and  regula- 
tion does  not  enter  into  the  question. 

5.  Switching  arrangements  are  very  simple. 

6.  Cost  of  right-of-way  will  usually  be  very  small  since  the 
line  may  be  placed  entirely  underground  making  it  possible  to 
cross  country  where  a  right-of-way  could  not  be  obtained.     In 
crossing  large  bodies  of  water  high-voltage  submarine  cables  may 
be  used  saving  the  expense  of  step-down  and  step-up  transformers 
required  in  a  similar  case  in  the  alternating-current  system.     To 
supply  electric-railway  substations  the  line  could  be  placed  under- 
ground along  the  right-of-way,  a  single  conductor  being  tapped 
at  each  substation. 

7.  A  single  conductor  with  earth  return  may  be  used  either 
as  the  ordinary  method  of  operation  or  in  case  of  emergency. 
All  other  grounds  must  then  be  removed  and  some  of  the  appar- 
atus insulated  for  the  full-line  voltage. 

Disadvantages. — 1.  Insulated  floors  and  couplings  are  required. 

2.  Units  are  of  moderate  size.     A  single  unit  delivering  250 
amp.  at  20,000  volts  has  a  rating  of  only  5,000  kw. 

3.  Line  loss  is  constant  independent  of  the  load. 

4.  Special  regulating  devices  are  required  to  keep  the  motor 
speed  constant. 

5.  Motors  have  no  overload  capacity,  since  the  current  is 
constant. 


CHAPTER  XV 


ELECTRICAL  INSTRUMENTS 

439.  Electrical  Instruments. — A  complete  study  of  electrical 
instruments  is  beyond  the  scope  of  this  book  but  in  the  following 
pages  the  principles  of  a  number  of  the  more  important  types 
used  in  the  measurement  of  electrical  quantities  are  discussed. 

440.  Direct-current  Voltmeters  and  Ammeters. — The  majority 
of  direct-current  voltmeters  and  ammeters  are  of  the  movable- 
coil  permanent-magnet  type  (Fig.  493).     A  light  rectangular  coil 
of  wire  M,  wound  on  an  aluminum  frame,  turns  in  the  field  of  a 
permanent  magnet  NS.     The  field  is  made  uniform  by  placing 
an  iron  cylinder  C  between  the  poles  and  this  gives  the  instru- 
ment a  uniform  scale.     The  coil  is  supported  on  jewelled  bearings 


Current  Terminal 


Current  Terminal 


FIG.  493. — Weston  perma- 
nent-magnet type  of  direct- 
current  voltmeter  or  ammeter. 


FIG.  494. — Ammeter 
shunt. 


FIG.  495. 


and  its  motion  is  controlled  by  two  spiral  springs  s,  s,  above  and 
below,  coiled  in  opposite  directions.  The  springs  are  used  to 
lead  the  current  in  and  out  of  the  coil.  The  aluminum  frame 
acts  as  a  damper  and  makes  the  instrument  dead  beat. 

A  large  resistance  r  must  be  inserted  in  series  with  the  coil  to 
limit  the  current  in  it  to  a  suitable  value.  The  resistance  re- 
quired varies  from  50  to  150  ohms  per  volt  and  the  current 
ranges  from  7  to  20  milliamp.  at  full-scale  deflection. 

441.  Voltmeter  Multipliers. — If  it  is  desired  to  change  the 
range  of  a  voltmeter  an  extra  resistance  rm,  Fig.  495,  must  be 
connected  in  series  with  the  movable  coil.  If  the  resistance  rm 

527 


528  ELECTRICAL  ENGINEERING 

of  the  multiplier  is  exactly  equal  to  the  resistance  of  the  coil  plus 
any  resistance  r  in  series  with  it,  then  for  a  given  line  voltage  the 
current  is  reduced  to  half  and  the  scale  reading  to  half.  Such 
a  multiplier  doubles  the  range  of  the  instrument  and  all  readings 
must  be  multiplied  by  2.  Multipliers  from  2  to  1  up  to  20  to  1 
are  used  with  direct-current  voltmeters. 

442.  Ammeter  Shunts. — The  ammeter  is  exactly  similar  to 
the  voltmeter  in  construction  (Fig.  493).     A  shunt  of  known  re- 
sistance ra,  Fig.  494,  is  placed  in  the  circuit  and  carries  the  cur- 
rent to  be  measured.     A  voltmeter  or  millivoltmeter  is  connected 
across  the  terminals  and  the  pointer  indicates  the  drop  of  voltage 
across  the  shunt.     This  drop  is  Irs  and  is  proportional  to  the 
current  since  the  resistance  ra  is  constant.     By  properly  arrang- 
ing the  resistance  of  the  shunt  and  the  resistance  r  in  series  with 
the  movable  coil  the  instrument  may  be  made  to  read  directly 
in  amperes. 

If  a  shunt  has  a  resistance  r8  =  0.001  ohm  and  carries  a  cur- 
rent of  100  amp.,  the  voltage  drop  is  0.1  volts  =  100  millivolts. 
Instead  of  calibrating  the  scale  in  millivolts  it  may  be  calibrated 
directly  in  amperes. 

If  the  ammeter  is  provided  with  a  number  of  shunts  it  may  be 
used  to  measure  a  large  range  of  currents.  For  the  smaller  cur- 
rent ranges  the  shunt  is  usually  placed  inside  the  meter  case. 

The  permanent  magnet  type  is  standard  for  direct-current  in- 
struments but  cannot  be  used  for  alternating  currents. 

443.  Thomson  Inclined-coil  Ammeter. — Fig.  496  shows  a  sec- 
tion of  an  inclined-coil  ammeter  or  voltmeter.     The  current  to 
be  measured  flows  through  the  inclined  coil  and  produces  a  mag- 
netic field  perpendicular  to  the  plane  of  the  coil.     A  soft-iron 
core  is  placed  in  the  field  and  tends  to  turn  until  it  becomes  par- 
allel to  the  direction  of  the  flux.     The  motion  is  controlled  by  a 
spiral  spring  and  the  counterweight  balances  the  moving  parts. 
The  motion  of  the  pointer  is  damped  by  a  light  aluminum  vane. 
In  the  ammeter  the  stationary  coil  consists  of  a  few  turns  of  large 
wire,  while  the  voltmeter  has  a  large  number  of  turns  of  small 
wire.     The  scale  is  not  uniform.     Inclined-coil  meters  may  be 
used  for  both  alternating  and  direct  currents  but  they  are  not  so 
good  as  the  permanent-magnet  direct-current  meters. 

444.  Weston  Soft-iron-type  Ammeters  and  Voltmeters. — This 
type  of  instrument  is  illustrated  in  Fig.  497.     A  is  a  fixed  tri- 
angular piece  of  soft  iron  bent  into  the  form  of  a  cylinder  and  B 


ELECTRICAL  INSTRUMENTS 


529 


is  a  movable  piece  carried  on  a  shaft.  A  magnetic  field  is  pro- 
duced by  the  coil  C  which  carries  the  current  to  be  measured  or  a 
current  proportional  to  the  voltage  to  be  measured.  The  two 
soft-iron  pieces  become  magnetized  in  the  same  direction  and 
repel  one  another,  producing  a  deflection  of  the  pointer.,  Motion 
is  controlled  by  a  spiral  spring  and  oscillations  are  damped  out 
by  a  vane  enclosed  in  an  air  chamber.  The  iron  pieces  are 
shaped  to  give  a  uniform  scale. 

Such  instruments  may  be  used  to  measure  direct  currents  as 
well  as  alternating  currents. 


FIG.  496. — Thomson  inclined- 
coil  ammeter. 


FIG.  497.— Weston  soft-  FIG.  498.— Elec- 
iron  type  ammeter  or  volt-  tro  dynamometer- 
meter,  type  voltmeter. 


445.  Electrodynamometer-type  Voltmeter. — The  electrodyna- 
rnometer-type  voltmeter,  Fig.  498,  depends  on  the  interaction 
between  two  coils  carrying  a  current  proportional  to  the  voltage 
to  be  measured.     The  permanent  magnet  of  Fig.  493  is  replaced 
by  a  coil  of  wire  in  series  with  the  moving  coil.     No  iron  is  used 
in  the  instrument.     This  construction  is  not  suitable  for  am- 
meters since  the  current  must  be  carried  by  a  moving  contact 
between  the  coils.     It  can  be  .used  for  voltmeters  but  other  types 
are  more  satisfactory.     Wattmeters  are,  however,  designed  on 
this  principle. 

446.  Hot-wire  Ammeters  and  Voltmeters. — The  expansion  of 
a  wire,  due  to  the  heat  produced  by  the  passage  of  current  through 
it,  is  utilized  in  this  type  of  meter  (Fig.  499).     AB  is  a  wire  of 
platinum-silver  alloy  which  carries  the  current  to  be  measured. 
As  it  expands  the  tension  on  the  fine  phosphor-bronze  wire  CD  is 
reduced  and  this  enables  the  spring  S  to  pull  the  silk  fiber  FE  to 
the  left.     The  silk  fiber  passes  around  a  pulley  on  the  shaft  of 
the  moving  system  and  causes  the  pointer  to  move  over  the 
scale.     The  motion  of  the  disc  is  damped  by  the  aluminum  disc 
which  turns  between  the  poles  of  the  permanent  magnet  shown 
in  Fig.  499 (a). 


530 


ELECTRICAL  ENGINEERING 


For  voltmeters  a  large  resistance  R  is  connected  in  series  with 
the  hot  wire  so  that  the  current  is  proportional  to  the  voltage  to 
be  measured.  The  zero  can  be  adjusted  by  the  screw  at  A 


n.A 


-Pointer 
Hot  Wire -Platinum  Siver 


^Phosphor  Br«nze  Wire 

Permanent 
agnet 


Terminal 


Series  Resistance 

W 

R 


Terminal 


FIG.  499. — Hot-wire  ammeter  or  voltmeter. 

447.  Dynamometer-type  Wattmeter. — The  dynamometer  prin- 
ciple is  satisfactory  when  applied  to  wattmeters,  Fig.  500,  because 
the  two  coils  are  not  connected  in  series  but  carry  different 
currents.  The  series  coil  of  large  wire  carries  the  line  current 
and  takes  the  place  of  the  permanent  magnet  in  the  direct- 
current  voltmeter,  Fig.  493.  The  voltage  coil  is  exactly  similar 
to  the  moving  coil  in  the  direct-current  voltmeter  and  is  supported 
and  controlled  in  the  same  way. 


(tt)    Uncompensated  (&)     Compensated 

FIG.  500. — Dynamometer  type  wattmeter. 

The  torque  exerted  on  the  moving  coil  is  proportional  to  the 
product  of  line  current  and  line  voltage,  that  is,  to  the  power  in 
the  circuit. 

If  necessary  a  shunt  may  be  used  in  parallel  with  the  current 
coil  and  a  multiplier  in  series  with  the  voltage  coil. 


ELECTRICAL  INSTRUMENTS 


531 


A  wattmeter  connected  as  in  Fig.  500  (a)  has  a  slight  zero  error, 
because  when  no  current  is  being  supplied  to  the  load,  the  current 
for  the  voltage  coil  passes  through  the  series  coil  and  the  watt- 


meter   indicates   a   small 


E2 

amount  of  power  =  Ei  =  -5-  where 

JK 


E  is  the  line  voltage  and  R  is  the  resistance  of  the  voltage  coil. 
To  correct  this  error  a  third  coil  C  called  the  compensating  coil 
Fig.  500(6)  is  added.  It  has  the  same  number  of  turns  as  the 
series  coil  but  is  wound  in  the  opposite  direction  and  its  m.m.f. 

E 

neutralizes  the  m.m.f.  of  the  small  current  i  =  -5" 

K 

For  ordinary  measurements  of  power  the  voltage  is  applied 
between  terminals  A  and  B,  but  when  the  meter  is  to  be  cali- 
brated, using  separate  voltage  and  current  sources,  no  zero  error 
is  introduced  and  a  third  terminal  D  must  be  provided  instead  of 
A.  In  series  with  it  is  a  resistance  rc  equal  to  the  resistance  of 
the  compensating  coil. 

Dynamometer-type  wattmeters  can  be  calibrated  with  direct 
current  and  used  on  alternating-current  circuits. 


Voltage  Coil  Compensating 

— O     E    o ^^      Coil 


Eesultant 
Flux 


(a) 


Current  Coil 

FIG.  501. — Induction- type  wattmeter. 

448.  Induction-type  Wattmeter. — An  induction  wattmeter  is 
shown  in  Fig.  501.  The  voltage  coils  VV  are  connected  in  series 
across  the  line  and  the  current  coils  are  connected  in  series  with 
the  line.  Both  sets  of  coils  are  wound  on  the  same  laminated 
iron  core. 

The  voltage  coils  on  the  two  legs  are  wound  in  the  same  direc- 
tion and  their  m.m.fs.  add.  The  circuit  is  highly  inductive  and 


532  ELECTRICAL  ENGINEERING 

the  current  through  it  is  nearly  90  degrees  behind  the  line  volt- 
age. The  flux  4>v  is  in  phase  with  the  current  and  is  therefore  in 
quadrature  behind  the  line  voltage. 

The  current  coils  on  the  two  legs  are  wound  in  opposite  direc- 
tions and  both  tend  to  produce  a  flux  4>c  in  the  gap  in  mechanical 
quadrature  with  the  flux  <j>v.  The  flux  <£c  is  in  phase  with  the  line 
current  and  with  non-inductive  load  it  is  90  degrees  ahead  of  the 
flux  0fl.  A  revolving  field  is  thus  produced  in  the  space  between 
the  poles. 

A  light  aluminum  cup  or  disc  D  is  pivoted  in  this  field  and 
tends  to  turn  with  it  but  is  opposed  by  spiral  springs.  The 
torque  is  proportional  to  the  product  of  the  voltage  and  the  line 
current,  that  is,  to  the  power  in  the  circuit. 

The  current  in  the  circuit  VV  and  the  flux  <frv  produced  by  it  are 
never  quite  90  degrees  behind  the  line  voltage  E  due  to  presence 
of  resistance  in  the  coil  and  to  iron  losses  in  the  core.  The  error 
introduced  would  be  negligible  for  power  factors  near  unity  but 
with  inductive  loads  it  would  be  serious.  To  make  the  potential 
flux  lag  exactly  90  degrees  behind  the  line  voltage  a  second  coil 
S  is  wound  on  the  upper  part  of  the  core  and  it  is  closed  through 
an  adjustable  resistance  r.  The  flux  </>v,  Fig.  501  (a),  induces  in 
the  coil  S  an  e.m.f.  and  current  approximately  90  degrees  behind 
it  and  this  current  produces  a  component  of  flux  <£s  in  phase  with 
itself.  <f>s  combines  with  <f>v  and  produces  a  resultant  potential 
flux  <f>  exactly  in  quadrature  behind  E.  <f>s  may  be  adjusted  by 
adjusting  the  resistance  r  in  series  with  the  coil  S. 

Polyphase  wattmeters  are  made  in  the  same  way  and  consist 
of  two  single-phase  elements  exerting  torque  on  a  single  disc. 

Induction-type  ammeters  and '  voltmeters  have  also  been 
constructed.  All  meters  of  this  type  are  accurate  only  for  the 
frequency  for  which  they  have  been  designed. 

If  the  controlling  springs  are  removed  from  a  wattmeter  and  a 
registering  mechanism  is  added  the  meter  becomes  a  watt-hour 
meter  and  registers  energy  consumption. 

449.  Power-factor  Meters. — Power-factor  meters  indicate 
the  power  factor  of  a  circuit;  they  are  constructed  on  the  same 
principles  as  wattmeters,  of  both  electrodynamometer  and  induc- 
tion types  and  may  be  either  single-phase  or  polyphase. 

The  circuits  of  a  Weston  single-phase  power-factor  meter  are 
shown  in  Fig.  502.  The  current  coil  F  is  similar  to  that  in  the 
wattmeter,  Fig.  500,  but  the  moving  system  is  made  up  of  two 


ELECTRICAL  INSTRUMENTS 


533 


similar  coils  C\  and  C2.  The  common  terminal  of  the  two  coils 
is  connected  to  one  side  of  the  line  and  the  other  terminals  are 
connected  through  a  resistance  R  and  an  inductance  L  to  the 
other  side  of  the  line.  The  current  in  Ci  is  therefore  in  phase 
with  the  line  voltage  and  the  current  in  €2  is  in  quadrature  behind 
the  line  voltage. 

At  non-inductive  load  the  currents  in  Ci  and  F  are  in  phase 
and  C\  will  be  forced  around  until  its  plane  is  parallel  to  the  plane 
of  F.  If  the  load  current  lags  90  degrees  behind  the  voltage  the 
current  in  C2  is  in  phase  with  that  in  F  and  C2  is  turned  into  the 
plane  of  F.  For  intermediate  power  factors  the  moving  element 
takes  up  intermediate  positions. 


TS/                    R 

V         r 

A  A   A   A   A 

ww 

\ 

"1 

Til 

/TTJOOOTN 

L 

% 

E3 

F 

C0 

FIG.  502. — Weston  single-phase  power-factor  motor. 


This  instrument  can  be  used  as  a  two-phase  power-factor  meter 
if  Ci  and  C2  are  each  connected  across  one  phase  in  series  with  a 
suitable  resistance. 

A  three-phase  meter  has  three  coils  on  the  moving  element 
connected  in  star  to  the  three  lines. 

450.  Frequency  Meters. — The  vibrating  reed  frequency  meter 
(Fig.  503)  contains  a  number  of  steel  strips  of  different  lengths 
fixed  at  one  end  and  free  to  vibrate  at  the  other.  An  electro- 
magnet is  placed  behind  the  strips  and  is  excited  from  the  circuit 
of  which  the  frequency  is  required.  The  strip  with  a  natural 
period  corresponding  to  that  of  the  magnetic  field  will  be  set 
in  vibration.  The  ends  of  the  reeds  are  painted  white  and  when 
one  is  in  vibration  it  shows  as  a  white  band.  The  periods  of  the 
reeds  are  adjusted  during  manufacture  by  attaching  minute 
weights  to  the  free  ends. 


534 


ELECTRICAL  ENGINEERING 


451.  The  Weston  Frequency  Meter.— The  elements  of  a  Wes- 
tern frequency  meter  are  shown  in  Fig.  504.  Two  stationary  coils 
A  and  B  are  fixed  at  right  angles ;  a  resistance  R  and  inductance 
X  are  connected  in  series  across  the  line ;  the  coil  A  is  connected 
in  series  with  an  inductance  XA  across  the  resistance  R  and  coil 
B  is  connected  in  series  with  a  resistance  RB  across  the  induc- 
tance X.  The  series  inductance  Xa  is  added  to  damp  out 
higher  harmonics. 


Alternating  Current 
Supply 


FIG.  503. — Vibrating  reed  frequency  meter. 


FIG.  504. — Weston  fre- 
quency meter. 


The  moving  element  consists  of  a  soft-iron  core  C  mounted  on 
a  shaft  without  control  of  any  kind  and  it  takes  up  a  position  in 
the  direction  of  the  resultant  of  the  two  fields. 

When  the  frequency  increases,  the  current  in  A  decreases  and 
that  in  B  increases  and  the  core  and  pointer  move  with  the 
resultant  field. 


1234          Incomiug  Machine 

FIG.  505. — Synchroscope. 

452.  Synchroscope. — A  synchroscope  indicates:  (1)  whether 
the  incoming  machine  is  running  too  fast  or  too  slow,  and  (2) 
the  exact  instant  when  synchronism  is  reached. 

One  form  of  synchroscope  is  shown  in  Fig.  505.  It  has  a  lam- 
inated bipolar  magnetic  circuit  M  excited  by  field  coils  FF,  which 
are  connected  across  the  alternating-current  busbars  at  1  and  2 
and  produce  an  alternating  field.  The  moving  core  C  is  also 
laminated  and  carries  two  windings  A  and  B  at  right  angles  to 


ELECTRICAL  INSTRUMENTS  535 

one  another.  Their  common  terminal  is  connected  to  one  side 
of  the  incoming  machine  at  4.  The  other  terminals  of  A  and  B 
are  connected  respectively  through  a  reactance  x  and  a  resistance 
r  to  the  other  side  of  the  machine  at  3. 

The  current  in  F  is  in  quadrature  behind  the  line  voltage,  the 
current  in  A  is  in  quadrature  behind  the  machine  voltage  and  the 
current  in  B  is  in  phase  with  the  machine,  voltage. 

When  the  incoming  machine  is  exactly  in  synchronism  the  coil 
A  takes  the  position  shown  in  the  figure  since  the  current  in  A 
is  in  phase  with  the  current  in  F.  When  the  machine  is  90  degrees 
behind  the  position  of  synchronism  the  current  in  B  is  in  phase 
with  the  current  in  F  and  the  armature  turns  through  90  degrees 
and  brings  the  coil  B  in  line  with  the  poles. 

For  intermediate  phase  relations  the  armature  takes  inter- 
mediate positions  such  that  the  revolving  field  produced  by  the 
armature  winding  is  in  line  with  the  field  poles  when  the  current 
in  F  is  maximum.  The  phase  relation  is  indicated  by  a  pointer 
on  the  dial  of  the  synchroscope. 

When  the  frequency  of  the  incoming  machine  is  lower  than 
that  of  the  line,  the  phase  of  the  current  in  A  continually  falls 
behind  that  of  F  and  the  pointer  rotates  in  the  direction  marked 
"Slow."  When  the  incoming  machine  is  running  too  fast,  the 
rotation  of  the  pointer  is  in  the  opposite  direction  marked  "  Fast." 

When  the  machine  is  running  exactly  at  synchronous  speed 
and  is  exactly  in  phase,  the  pointer  is  vertical  and  stationary. 
The  switch  can  then  be  closed  and  the  speed  and  excitation  ad- 
justed until  the  machine  takes  its  proper  share  of  the  load  and 
operates  at  the  proper  power  factor.  . 

The  synchroscope  described  is  for  a  single-phase  circuit.  It 
can  be  used  for  a  two-phase  machine  by  connecting  the  two  coils 
A  and  B  to  the  two  phases  of  the  machine  and  the  coils  FF  across 
one  phase  of  the  line. 

For  three-phase  systems  the  armature  carries  a  three-phase 
winding  connected  to  the  three  phases  of  the  machine. 

453.  Tirrill  Regulator. — The  Tirrill  regulator  is  an  automatic 
voltage  regulator  designed  to  maintain  a  steady  voltage  at  the 
terminals  of  a  direct-current  generator  irrespective  of  ordinary 
load  fluctuations  or  changes  in  generator  speed.  It  can  also  be 
made  to  compensate  for  line  drop  by  increasing  the  generator 
voltage  as  the  load  increases. 

The  regulator  controls  the  voltage  by  rapidly  opening  and  clos- 


536 


ELECTRICAL  ENGINEERING 


ing  a  shunt  circuit  across  the  field  rheostat  of  the  generator.  The 
rheostat  is  so  adjusted  that  when  in  circuit  it  tends  to  reduce  the 
voltage  considerably  below  normal  and  when  short-circuited  the 
voltage  tends  to  rise  above  normal.  The  relative  lengths  of 
time  during  which  the  short-circuit  is  closed  or  opened  determines 
the  average  value  of  the  field  current  and  therefore  the  value  of 
the  terminal  voltage. 

The  method  of  operation  of  the  regulator  is  illustrated  in  Fig. 
506.  The  regulator  consists  essentially  of  two  magnets  control- 
ling two  sets  of  contacts.  The  main  control  magnet  has  two  in- 
dependent windings,  one,  the  potential  winding,  connected  across 


Compensating  Resistance  Shunt 

'O 


Potential  Winding 


Compensating  Winding 
Main  Control  Magnet 


FIG.  506. — Automatic  voltage  regulator. 

the  generator  terminals  and  the  other  across  a  shunt  in  the  load 
circuit.  The  latter  is  the  compensating  winding  and  is  used  only 
when  a  rise  of  voltage  with  load  is  required.  The  relay  magnet 
is  differentially  wound  and  controls  the  circuit  shunting  the  field 
rheostat.  The  operation  is  as  follows:  When  the  short-circuit 
across  the  field  rheostat  is  opened  the  voltage  tends  to  fall  below 
normal.  The  main  control  magnet  is  weakened  and  allows  the 
spring  to  pull  out  the  movable  core  until  the  main  contacts  are 
closed.  This  closes  the  second  circuit  of  the  differential  relay 
and  demagnetizes  it.  The  relay  spring  then  lifts  the  armature 
and  closes  the  relay  contacts.  The  field  rheostat  is  short-cir- 
cuited and  the  field  current  and  terminal  voltage  tend  to  rise. 
The  main  control  magnet  is  strengthened  and  opens  the  main 
contacts  allowing  the  differential  relay  to  open  the  short-circuit 
across  the  field  rheostat.  The  terminal  voltage  falls  again  and 


ELECTRICAL  INSTRUMENTS 


537 


this  cycle  of  operations  is  repeated  at  a  very  rapid  rate  main- 
taining a  steady  voltage  at  the  generator  terminals.  When  the 
compensating  winding  is  not  used  the  terminal  voltage  is  main- 
tained constant. 

When  it  is  necessary  to  compensate  for  line  drop  and  maintain 
a  constant  voltage  at  the  receiver  end  of  the  line,  the  compen- 
sating winding  is  connected  across  a  shunt  in  the  load  circuit. 
The  resistance  of  the  shunt  is  adjusted  to  give  the  required  com- 
pounding. The  compensating  winding  opposes  the  action  of  the 
potential  winding  on  the  main  control  magnet  so  that  as  load 
increases  a  higher  potential  is  necessary  at  the  generator  ter- 
minals in  order  to  close  the  main  contacts  and  open  the  shunt 
across  the  field  rheostat.  Thus,  the  generator  voltage  rises  with 
load.  The  condenser  connected  across  the  relay  contacts  serves 
to  reduce  the  sparking  when  the  circuit  is  opened. 


Main 


Contacts 


Compensating 
Winding^ 


A.C.  Field  A.C.  Generator 

Bheostat 
FIG.  507. — Automatic  voltage  regulator  for  alternating-current  generators. 

454.  Automatic  Voltage  Regulator  for  Alternating-current 
Generators. — The  regulator  described  in  Art.  453  may  be  modi- 
fied to  regulate  the  voltage  of  alternators.  The  desired  voltage 
is  maintained  by  opening  and  closing  a  short-circuit  across  the 
exciter  field  rheostat. 

The  method  of  operation  of  the  regulator  can  be  understood 
from  the  diagram  of  connections  shown  in  Fig.  507.  The  direct- 
current  control  magnet  is  connected  to  the  exciter  busbars  and 
has  a  fixed  core  in  the  bottom  and  a  movable  core  in  the  top 
attached  to  a  pivoted  lever,  at  the  other  end  of  which  is  a  spring 


538  ELECTRICAL  ENGINEERING 

which  closes  the  main  contacts.  The  alternating-current  control 
.magnet  has  a  potential  winding  connected  across  one  phase  of  the 
alternator  and  it  may  also  have  a  compensating  winding  con- 
nected through  a  current  transformer  to  one  of  the  feeders.  The 
core  is  movable  and  is  connected  to  a  pivoted  lever  controlled  by 
a  counterweight.  The  relay  magnet  is  differentially  wound  and 
is  connected  as  shown. 

Operation. — The  direct-  and  alternating-current  control  mag- 
nets are  adjusted  for  the  required  voltage  by  means  of  the  coun- 
terweight. The  exciter  field  rheostat  is  then  set  to  reduce  the 
voltage  about  65  per  cent,  below  normal.  This  weakens  both  of 
the  control  magnets  and  the  spring  closes  the  main  contacts  and 
demagnetizes  the  relay  magnet.  The  pivoted  armature  is  re- 
leased and  the  relay  contacts  are  closed  and  thus  short-circuit 
the  exciter  field  rheostat  and  immediately  raise  the  exciter  voltage 
and  the  alternator  voltage.  When  the  alternator  voltage  reaches 
the  value  for  which  the  regulator  is  adjusted,  the  main  contacts 
open  again,  the  relay  magnet  is  again  magnetized  and  the  short- 
circuit  on  the  exciter  field  rheostat  is  opened.  This  reduces  the 
voltage  as  before  and  the  cycle  of  operations  is  repeated  at  a 
very  rapid  rate  and  maintains  a  constant  voltage  at  the  terminals 
of  the  alternator  if  the  compensating  winding  is  not  connected. 


Three-phase  Generator 


LJ,   rr^ri 


1  ..... 


(  a )  .S.ingle-phase  (  ft  }  Three.phase 

FIQ.  508. — Line  compensators. 

When  it  is  necessary  to  maintain  a  constant  voltage  at  the 
receiver  end  of  the  line  the  compensating  winding  is  connected  as 
shown.  As  the  load  increases  it  brings  the  main  contacts  closer 
together  and  so  increases  the  time  of  short-circuit  of  the  field 
rheostat  and  thus  increases  the  terminal  voltage  of  the  alternator. 
Using  a  current  transformer  and  a  dial  switch  any  line  drop  up  to 
15  per  cent,  can  be  compensated  for,  but  only  at  a  given  power 
factor.  When  the  power  factor  of  the  load  varies  through  a 
wide  range  a  line  compensator,  Fig.  508 (a),  should  be  used  in  con- 


ELECTRICAL  INSTRUMENTS  539 

junction  with  the  potential  coil  and  the  compensating  coil  should 
be  disconnected. 

The  line  compensator  forms  a  local  circuit  with  the  same  vol- 
tage characteristics  as  the  main  line.  The  shunt  transformer  TI 
gives  a  secondary  voltage  proportional  to  the  generator  voltage. 
The  current  transformer  T2  produces  through  the  circuit  rx  a  cur- 
rent proportional  to  the  load  current,  r  is  a  resistance  which 
consumes  a  voltage  proportional  to  and  in  phase  with  the  resist- 
ance drop  in  the  line  and  this  voltage  is  transferred  to  the  com- 
pensator circuit  by  the  potential  transformer  TV  #  is  a  reactance 
and  consumes  a  voltage  proportional  to  the  reactance  drop  in 
the  line.  This  voltage  is  transferred  to  the  compensator  circuit 
by  the  transformer  T*  which  also  forms  the  reactance.  Thus 
there  are  in  the  compensator  circuit  three  voltages  proportional 
respectively  to  the  generator  voltage  and  the  resistance  and 
reactance  drops.  If  the  same  proportions  are  maintained  in  each 
case,  the  voltage  between  the  terminals  AB  will  be  proportional 
to  the  voltage  at  the  end  of  the  line,  and,  therefore,  if  the  poten- 
tial coil  of  the  regulator  is  connected  across  AB  it  will  maintain 
a  constant  voltage  at  the  receiver  end  of  the  line.  In  the  case 
of  transmission  lines  of  very  high  voltage  a  correction  must  be 
made  for  the  capacity  of  the  lines.  With  a  three-phase  alter- 
nator, as  in  Fig.  507,  two  current  transformers  must  be  used 
connected  as  shown  in  Fig.  508(6). 

One  automatic  voltage  regulator  can  be  applied  to  control  the 
voltage  of  a  system  operating  two  or  more  alternators  in  parallel. 

The  regulator  may  also  be  applied  to  the  exciter  of  a  syn- 
chronous-phase modifier  to  maintain  a  constant  voltage  at  the 
receiver  end  of  the  line. 


INDEX 


Absampere,  definition  of,  54 
Abvolt,  definition  of,  52 
Adjustable  speed  operation,  210 
Admittance,  131,  132,  135 

stator  exciting,  465 
Air-blast  transformers,  386 
All-day  efficiency,  382 
Alternator  windings,  289 
Alternators,  rated  speed  of,  355 
rating  of,  300 
types  of,  280 

Aluminum,  properties  of,  83 
Alternating-current    commutator 

motor  with  shunt  character- 
istics, 505 
series  motor,  494 

speed  control  of,  500 
Ammeters,  527 
Ampere,  definition  of,  54 

turn,  61 

Analysis  of  complex  waves,  146 
Argon  filled  rectifier,  442 
Armature  copper  loss,  216 
core,  165 
reaction    in  alternating-current 

generators,  302 
in  direct-current   generators, 

182 

in  polyphase  generators,  305 
in  single-phase  generators,  309 
in    synchronous    converters, 

426 

resistance,  275,  312 
windings,  167 

Asynchronous  phase  modifier,  488 
Automatic  voltage  regulator,  535 
Auto-transformer,  396 


B 


Balancers,  249 


Bearing  friction  loss,  220 

Blondel    diagram    for    synchronous 

motor,  334,  339 
Boosters,  247 

negative,  249 

series,  248 

Booster  transformers,  396 
British  thermal  unit,  80 
Brush  contact  drop,  216 

contact  loss,  216 

effect  of  moving  the,  181 

friction  loss,  220 

lifting  device,  434 
Brushes,  178,  262 
Bucking,  435 
Burning  of  brushes,  236 


Calorie,  80 
Capacity,  12 

in  alternating-current  circuits, 
111 

of  concentric  cylinders,  15 

of  single  conductor  cable,  16 

of  parallel  conductors,  17,  19 

unit  of,  13 

Carter's  fringing  constant,  269 
Cascade     operation     of     induction 

motors,  476 
Charging    current    of    transmission 

line,  509 
Circle  diagram  of  induction  motor, 

458,  460,  461 
Coercive  force,  71 
Commutating  e.m.f.,  230,  231 
Commutation,  224,  253 

in    alternating-current    series 
motor,  499 

in  repulsion  motor,  503 

time  of,  225 
Commutator,  176 

peripheral  speed  of,  262 


541 


542 


INDEX 


Complex  alternating  waves,  142 
analysis  of,  146 
average  value  of,  142 
effective  value  of,  142 
form  factor  of,  142 
Compensating  winding,  242,  498 
Compensators  for  three-wire  genera- 
tors, 437-8 

Compound  excitation,  179 
Compounding  converters,  430 
curve  of  alternator,  317 
direct-current  generator,  190 
synchronous  motor,  330,  342 
Concatenation  of  induction  motors, 

476 

differential,  478 
Condenser,  14 
bushing,  40 
energy  stored  in,  25 
Condensers  in  multiple,  24 

series,  24 

Conductance,  81,  131,  134 
Conductivity,  81 

standard  of,  81 
Conductors,  1 

Constant  current  transformer,  401 
potential  to  constant  current, 

140 

transformer,  363 
Converters,  frequency,  414,  493 
phase,  414,  491 
split-pole,  432 
synchronous,  414 

booster,  431 

Cooling  of  transformers,  384 
Copper  losses,  215 

properties  of,  83 
Core  loss,  222 

current,  364 
type  transformer,  383 
Corona,  45 

Coulomb,  definition  of,  2 
Cross-magnetizing,  185,  187,  275 
Current  capacity  of  wires,  85,  87 
densities  in  alternators,  356 
brush  contacts,  236 
direct-current  armatures,  261 
field  windings,  274 
transformer,  phase  angle  of,  400 


Current    transformer,   ratio  correc- 
tion factor  of,  400 
unit  of,  54 
Cylindrical  rotors,  282,  361 


D 


Damping  grids,  343,.  436 
Demagnetizing  m.m.f.,  185,  187,  275 
Design  of  alternators,  354 

direct-current  machines,  264 

transformers,  405 
Dielectric  constant,  2,  43 

flux,  4 

hysteresis,  44 

losses,  44 

permeance,  13 

strength,  41-43 
Dispersion  coefficient,  263 
Distribution  factors,  290 
Double  current  generators,  436 

windings,  174 
Drum  winding,  168-172 


E 


Eddy-current  loss  in  armature  cop- 
per, 221 

in  direct'current  machines,  218 

in  transformers,  379 
Effective  value  of  a  sine  wave,  106 

complex  wave,  142 
Efficiency     of     direct-current    ma- 
chines, 222,  252 

transformers,  381,  382 
Electric  energy,  55 

loading,  259,  261 

power,  55 
Electrical    measuring    instruments, 

527 

Electrification,  1 

Electrodynamometer  voltmeter,  529 
Electromagnetics,  2,  51 
Electromotive  force  equation  of  an 
alternator,  284,  300 

a  direct-current  generator,  180 

unit  of,  52 
Electrostatic  field,  3,  4,  5 

stresses  in,  27 


INDEX 


543 


Electrostatics,  laws  of,  2 
Equalizer  connection,  242 

rings,  171 

Equipotential  surfaces,  9,  21 
Equivalent  sine  wave,  142 
Excitation  of  alternators,  358 

regulation,  322 

Exciting  current  of  induction  motor, 
478  :    v" 

transformer,  368 


Farad,  definition  of,  13 
Field  characteristic  of  direct-current 
generator,  190 

windings,  178 
Flashing,  239 
Form  factor  of  alternating  waves, 

142,  286 

Fourier  series,  146 
Frequency  converter,  493 

meters,  533 

of  hunting,  352 


G 


Induced  charges,  6 
Inductance,  92,  107 

of  armature  coil,  227 
of  iron-clad  circuits,  94 
Induction  frequency  converter,  493 
generator,  486 
motor,  444 

analysis   by   rectangular    co- 
ordinates, 465 
applications  of,  475 
circle  diagram  of,  454 
single-phase,  482 
speed  control  of,  476 
starting,  473,  481 
synchronous  speed,  447 
regulator,  403 
type  wattmeters,  531 
Inductive  compensation,  498 
Inductor  alternator,  282 
Insulation,  breakdown  of,  43 
of  alternators,  356 
resistance,  44 
Insulators,  1 
Interpole  motors,  213 
Interpoles,  237 


Graded  insulation  for  cables,  35 

H 

Harmonics  in  e.m.f.  wave,  296,  297 
Heat  units,  80 
Heating,  253 

of  converters,  422 
Henry,  definition  of,  93 
High  voltage  direct-current  system 

of  transmission,  524 
Homopolar  generators,  251 
Hot  cathode  rectifier,  442 

wire  instruments,  529 
Hottest  spot  temperatures,  255 
Hunting,  frequency  of,  352 

of  synchronous  machines,  351 
Hysteresis,  71 

loss,  271,  378 
Hysteretic  constants,  73 


Joule,  definition  of,  56 
Joule's  law,  79 

K 

Kirchoff's  laws,  88 

applied    to    alternating-current 
circuits,  141 


Lead,  properties  of,  83 
Leakage  fluxes,  304 

reactances  of  induction  motors, 

479 

of  transformers,  368 
Lifting  magnets,  76 
Limits  of  output   of   direct-current 
machines,  252 


544 


INDEX 


Line  compensator,  538 

Load  characteristics  of  synchronous 

motors,  331,  340 
Losses  in  alternators,  359 

direct-current  machines,  215 

transformers,  378 


Open-circuit    test    of    transformers, 

374 

-delta  connection,  391 
Output  equation  of  alternating-cur- 
rent generator,  359 
direct-current  generator,  259 


M 

Magnetic  characteristics,  69 
of  cast  iron,  70 
of  cast  steel,  70 

field,  47,  67 

energy  stored  in,  66 

flux,  47 

leakage,  263 

loading,  259 

materials,  73 

potential,  48 
Magnetism,  laws  of,  47 

theories  of,  75 
Magnetization,  46 

curves  for  cast  iron,  271 
,cast  steel,  271 
sheet  steel,  271 
Magnetizing  current,  364 

force,  47 
Magnetomotive  force,  49 

unit  of,  61 

Manganese,  properties  of,  83 
Maxwell's  corkscrew  rule,  51 
Mercury,  properties  of,  83 

vapor  rectifiers,  440 
Motor  starter,  209 
Multiple-circuit  windings,  294 

wire  systems,  211 
Mutual  inductance,  95 

N 

N-phase  converter,  419 
No  load  neutral,  183 
No-voltage  release,  209 


Ohm,  definition  of,  79 
Ohm's  law,  79 


Parallel  operation  of  alternators,  346 

direct-current   generators,    242 

synchronous  converters,  435 

three-wire  generators,  439 

transformers,  395 
Permeability,  50 
Permeance,  dielectric,  13 

magnetic,  51 

Phase    angle   of   instrument   trans- 
formers, 398,  400 

characteristics   of   synchronous 
motor,  333,  341 

converter,  491 

splitting,  485 

Platinum,  properties  of,  83 
Pole  face  loss,  220 

pieces,  165 

Polyphase    alternating-current    cir- 
cuits, 152 

armature  reaction,  305 

commutator  motor,  501 
Potential,  5 

gradient,  9-11,  16,  19 

transformer,  397 
Potentiometer,  91 
Power,  electric,  55 

-factor,  116,  121 
meters,  532 

in   alternating-current  circuits, 
116 

in  three-phase  circuits,  158 
measurement  of,  159 

units  of,  56 
Progressive  winding,  174 


R 


Kadiator  tank,  386 

Ratio  correction  factors  for  instru- 
ment transformers,  398, 
400 


INDEX 


545 


Reactance,  50,  131,  132 

condensive,  111 

inductive,  109 

of  transformers,  408 

voltage,  228,  233 
Reactive  factor,  121 
Rectangular  coordinates,  134 
Rectifier,  hot  cathode,  argon  filled, 
442 

mercury  vapor,  440 
Regulation,  252 

curves  of  alternators,  317 

direct-current  generators,  188, 

196 

transformers,  376 
Repulsion  motor,  501 

compensated,  504 
Resistance,  79,  80 

commutation,  230 

temperature  coefficient  of,  81 
Resistances  in  parallel,  90 

series,  90 

Resonant  circuit,  128 
Reversing  induction  motor,  450 
Revolving  field,  307 
Rosenberg  generator,  250 
Rotor,  450,  475 


Saturation  curve,  no  load,  186,  270 

under  load,  187 
Scott  connection,  393 
Self -inductance,  95 
Self-starting    synchronous    motors, 

343 

Semi-enclosed  motors,  256 
Separate  excitation,  179 
Series  excitation,  179 
Series  field-copper  loss,  215 
Series-parallel  speed  control,  212 
Shading  coils,  485 
Shell-type  transformers,  383 
Short-circuit  currents  of  alternators, 

323 
test  of  transformers,  375 

pitch  windings,  171,  235,  294 
Shunt  excitation,  179 

-field  copper  loss,  215 

35 


Silicon  steel,  74 
Silver,  properties  of,  83 
Sine  wave,  104 

average  value  of,  106 

effective  value  of,  106 
Single-phase  induction  motor,  482 
Slip,  450 

Spacing  of  conductors,  510 
Sparkless  commutation,  233 
Specific  inductive  capacity,  2 

resistance,  81 

Speed  characteristics  of  motors,  201 
construction  of,  .204 

equation       of       direct-current 
motor,  199 

methods  of  varying,  199 

variation  with  line  voltage,  206 
Split-pole  converter,  432 
Starter    for    direct-current    motor, 

209 
Starting  induction  motors,  473,  481 

synchronous  converters,  433 
motors,  342 

torque,     direct-current    motor, 

208 

induction  motor,  469 
Stator  of  induction  motor,  444 
Storage  batteries,  245 
Surface  leakage,  45 
Synchronizing  power,  338 
Synchronous  booster  converter,  431 

motor,  327 

phase  modifier,  345 

speed,  328,  447 
Synchroscope,  534 


Tee  connection,  392 
Temperature  limits,  253 
Third  harmonics  in  alternators,  299 
Thompson    inclined    coil    ammeter, 

528 

Three-wire  generators,  437 
Thury  system,  524 
Time  constant  of  a  circuit,  99 
Tirrill  regulator,  535 
Tooth  taper,  268 
Torque  characteristics  of  motors,  204 


546 


INDEX 


Torque,    characteristics    of    motors, 
construction  of,  207 

starting,  208,  469 
Totally  enclosed  motors,  268 
Transformer,  363 

constant  current,  401 

constants  of  a,  374 
Transmission  systems,  507 
Tubular  tank,  385 
Tungsten,  properties  of,  83 
Turbo-alternators,  284 


"V"  curves  for  synchronous  motor, 

341 
Vector  diagrams  for  alternators,  313 

synchronous  motors,  328 

transformers,  366 
Vent  ducts,  167 
Ventilation,  359 
Volt,  definition  of,  6 
Voltage    characteristics  of  alterna- 
tors, 315 


Voltage    characteristics    of    direct- 
current  generators,   188-193 
transformers,  378 

Voltmeters,  527 

W 

Ward  Leonard  system,  211 
Watt,  definition  of,  56 

-ratio  curve,  161 
Wattmeter,  dynamometer  type,  530 

induction  type,  531 
Weston  soft  iron  instruments,  528 
Windage  loss,  220 
Wire  table,  85 
Wound  rotor,  474 


"Y"  connection,  155 
Yoke,  165 


Zig-zag  leakage  flux,  479 


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